Explanation:
Okay, so I need to find the number of quadratic equations of the form $x^2 + ax + b = 0$ such that whenever $c$ is a root, $c^2 - 2$ is also a root. Hmm, let me think about how to approach this.

First, let me recall that a quadratic equation has two roots (counting multiplicity). Let me denote the roots as $r_1$ and $r_2$. According to Vieta's formulas, we have:

$r_1 + r_2 = -a$ and $r_1 r_2 = b$.

Now, the condition given is that if $c$ is a root, then $c^2 - 2$ is also a root. So, for each root $c$, $c^2 - 2$ must be another root. Since there are only two roots, this condition must hold for both roots. Let me consider the possibilities.

Case 1: The quadratic has a repeated root. That is, $r_1 = r_2$. Then, if $c = r_1$, then $c^2 - 2$ must also be a root. But since there's only one root, $c^2 - 2$ must equal $c$. So, solving $c^2 - 2 = c$, which gives $c^2 - c - 2 = 0$. Solving this quadratic equation:

$c = [1 \pm \sqrt{1 + 8}]/2 = [1 \pm 3]/2$, so $c = 2$ or $c = -1$.

Therefore, if the quadratic has a repeated root, it must be either 2 or -1. Let me check if these satisfy the condition.

If the root is 2, then $c^2 - 2 = 4 - 2 = 2$, which is the same root. So that works. Similarly, if the root is -1, then $c^2 - 2 = 1 - 2 = -1$, which is also the same root. So both 2 and -1 are valid repeated roots. Therefore, there are two quadratics in this case: $(x - 2)^2 = x^2 - 4x + 4$ and $(x + 1)^2 = x^2 + 2x + 1$. So that's two equations.

Case 2: The quadratic has two distinct roots, $r_1$ and $r_2$. Then, the condition implies that if $r_1$ is a root, then $r_1^2 - 2$ must be a root. Similarly, if $r_2$ is a root, then $r_2^2 - 2$ must be a root. Since there are only two roots, this means that $r_1^2 - 2$ must be either $r_1$ or $r_2$, and similarly for $r_2^2 - 2$.

So, let me consider the possibilities for each root:

For root $r_1$:

Either $r_1^2 - 2 = r_1$ or $r_1^2 - 2 = r_2$.

Similarly for root $r_2$:

Either $r_2^2 - 2 = r_2$ or $r_2^2 - 2 = r_1$.

But since there are two roots, we need to consider combinations of these possibilities. Let me think about possible scenarios.

Subcase 2a: Both roots satisfy $c^2 - 2 = c$. Then, both roots are solutions to $c^2 - c - 2 = 0$, which we already solved earlier, giving roots 2 and -1. But if both roots are 2 and -1, then the quadratic would be $(x - 2)(x + 1) = x^2 - x - 2$. However, this is a quadratic with distinct roots. Wait, but in this case, the roots are 2 and -1. Let me check if this satisfies the condition. If c is 2, then $c^2 - 2 = 4 - 2 = 2$, which is already a root. If c is -1, then $c^2 - 2 = 1 - 2 = -1$, which is also a root. So this quadratic does satisfy the condition. However, this quadratic was not considered in Case 1 because it has distinct roots. Wait, but in Case 1, we considered repeated roots. So this quadratic is a separate case. Therefore, this is another quadratic. But wait, but in this case, both roots satisfy $c^2 - 2 = c$, so this is a valid quadratic. However, this quadratic is different from the ones in Case 1. So, this is another quadratic. Wait, but hold on, if both roots satisfy $c^2 - 2 = c$, then the quadratic equation would have roots 2 and -1. So that's one quadratic. But is there another quadratic in this subcase? No, because the equation $c^2 - c - 2 = 0$ has exactly two roots, so the quadratic with roots 2 and -1 is unique. Therefore, this is one quadratic in Subcase 2a.

Subcase 2b: One root satisfies $c^2 - 2 = c$, and the other root satisfies $c^2 - 2$ equals the other root. Wait, let me think again. Let me consider that for each root, $c^2 - 2$ must be another root. So, if one root is 2 (which satisfies $c^2 - 2 = c$), then the other root must satisfy that if you take $c^2 - 2$, it must be either 2 or the other root. Similarly for the other root.

Alternatively, maybe there's a better way to approach this. Let me think.

Suppose that the roots are $r_1$ and $r_2$. Then, the condition implies that:

If $r_1$ is a root, then $r_1^2 - 2$ is also a root. Therefore, $r_1^2 - 2$ must be either $r_1$ or $r_2$. Similarly, $r_2^2 - 2$ must be either $r_1$ or $r_2$.

So, there are several possibilities here. Let me consider different possibilities:

Possibility 1: $r_1^2 - 2 = r_1$ and $r_2^2 - 2 = r_2$. Then, both roots satisfy $c^2 - c - 2 = 0$, so they are 2 and -1. This is the quadratic we considered in Subcase 2a.

Possibility 2: $r_1^2 - 2 = r_2$ and $r_2^2 - 2 = r_1$. So, this is a system of equations. Let me write that down:

$r_1^2 - 2 = r_2$ ...(1)

$r_2^2 - 2 = r_1$ ...(2)

So, substituting equation (1) into equation (2):

From (1): $r_2 = r_1^2 - 2$. Substitute into (2):

$(r_1^2 - 2)^2 - 2 = r_1$

Expanding that:

$r_1^4 - 4r_1^2 + 4 - 2 = r_1$

Simplify:

$r_1^4 - 4r_1^2 + 2 = r_1$

Bring all terms to one side:

$r_1^4 - 4r_1^2 - r_1 + 2 = 0$

Hmm, quartic equation. Let me try to factor this. Maybe there are rational roots. Let me apply Rational Root Theorem. Possible rational roots are ±1, ±2.

Testing r = 1: 1 - 4 - 1 + 2 = -2 ≠ 0

Testing r = -1: 1 - 4 + 1 + 2 = 0. Oh, r = -1 is a root.

So, we can factor (r + 1) out of the quartic.

Let me perform polynomial division or use synthetic division.

Dividing $r^4 - 4r^2 - r + 2$ by (r + 1):

Using synthetic division for root -1:

Coefficients: 1 (r^4), 0 (r^3), -4 (r^2), -1 (r), 2 (constant)

Bring down the 1.

Multiply by -1: -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: 1. Add to next coefficient: -4 + 1 = -3.

Multiply by -1: 3. Add to next coefficient: -1 + 3 = 2.

Multiply by -1: -2. Add to last coefficient: 2 + (-2) = 0. Perfect.

So, the quartic factors as (r + 1)(r^3 - r^2 - 3r + 2).

Now, let's factor the cubic: $r^3 - r^2 - 3r + 2$.

Again, try rational roots: ±1, ±2.

Testing r = 1: 1 - 1 - 3 + 2 = -1 ≠ 0.

Testing r = 2: 8 - 4 - 6 + 2 = 0. Yes, r = 2 is a root.

Divide the cubic by (r - 2):

Using synthetic division:

Coefficients: 1 (r^3), -1 (r^2), -3 (r), 2 (constant)

Root at 2:

Bring down 1.

Multiply by 2: 2. Add to next coefficient: -1 + 2 = 1.

Multiply by 2: 2. Add to next coefficient: -3 + 2 = -1.

Multiply by 2: -2. Add to last coefficient: 2 + (-2) = 0.

So, cubic factors as (r - 2)(r^2 + r - 1).

Therefore, the quartic factors as (r + 1)(r - 2)(r^2 + r - 1).

Therefore, the roots of the quartic equation are r = -1, r = 2, and roots of $r^2 + r - 1 = 0$, which are $r = [-1 \pm \sqrt{1 + 4}]/2 = [-1 \pm \sqrt{5}]/2$.

So, the solutions to the quartic equation are r = -1, 2, and $[-1 \pm \sqrt{5}]/2$.

Now, going back to our system of equations (1) and (2):

We have that $r_1$ and $r_2$ satisfy $r_1^2 - 2 = r_2$ and $r_2^2 - 2 = r_1$. So, the solutions for $r_1$ and $r_2$ can be:

Either:

- $r_1 = -1$, then from equation (1): $r_2 = (-1)^2 - 2 = 1 - 2 = -1$. So, both roots are -1. But this is a repeated root, which was considered in Case 1. However, in this case, since we are in Subcase 2b where we are considering distinct roots? Wait, no. Wait, in Subcase 2b, we are considering the case where one root satisfies $c^2 - 2 = c$ and the other satisfies $c^2 - 2$ equals the other root. Wait, but if both roots are -1, then they are the same root, so that's a repeated root. But in this case, we are in Case 2 where we have two distinct roots. Therefore, this solution (both roots -1) is actually covered in Case 1, so we need to exclude it here.

Similarly, if $r_1 = 2$, then from equation (1): $r_2 = 2^2 - 2 = 4 - 2 = 2$. Again, same root, so repeated root. So again, covered in Case 1.

Therefore, the other roots are $r = [-1 \pm \sqrt{5}]/2$. Let me denote them as $r_1 = [-1 + \sqrt{5}]/2$ and $r_2 = [-1 - \sqrt{5}]/2$. Let me check if they satisfy the equations.

First, check if $r_1^2 - 2 = r_2$:

Compute $r_1^2$:

Let me compute $r_1 = (-1 + √5)/2$. Then,

$r_1^2 = [(-1)^2 + 2*(-1)*(√5) + (√5)^2]/4 = [1 - 2√5 + 5]/4 = (6 - 2√5)/4 = (3 - √5)/2.

Then, $r_1^2 - 2 = (3 - √5)/2 - 2 = (3 - √5 - 4)/2 = (-1 - √5)/2 = r_2$. Perfect, that's correct.

Similarly, check $r_2^2 - 2 = r_1$:

Compute $r_2 = (-1 - √5)/2$. Then,

$r_2^2 = [(-1)^2 + 2*(-1)*(-√5) + (√5)^2]/4 = [1 + 2√5 + 5]/4 = (6 + 2√5)/4 = (3 + √5)/2.

Then, $r_2^2 - 2 = (3 + √5)/2 - 2 = (3 + √5 - 4)/2 = (-1 + √5)/2 = r_1$. Correct.

Therefore, these two roots satisfy the condition. Therefore, the quadratic equation with roots $r_1$ and $r_2$ is:

$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$.

Compute $r_1 + r_2$:

$[(-1 + √5)/2 + (-1 - √5)/2] = (-2)/2 = -1$.

Compute $r_1 r_2$:

Multiply the two roots:

$[(-1 + √5)/2] * [(-1 - √5)/2] = [(-1)^2 - (√5)^2]/4 = [1 - 5]/4 = (-4)/4 = -1$.

Therefore, the quadratic equation is $x^2 - (-1)x + (-1) = x^2 + x - 1 = 0$. So that's another quadratic equation.

Therefore, in this Subcase 2b, we have one quadratic equation: $x^2 + x - 1 = 0$.

Wait, but hold on, are there other possibilities? For example, if we take $r_1 = [-1 + √5]/2$ and $r_2 = [-1 - √5]/2$, which gives us the equation above. But since the roots are distinct, this is one quadratic. However, could there be another quadratic if we take different roots? For example, if we take $r_1 = [-1 + √5]/2$ and $r_2 = something else? Wait, no, because the system of equations only gives us these roots. So, only one quadratic here.

But wait, let me check if there are other solutions. The quartic equation had roots -1, 2, and the two roots from the quadratic. But we already considered -1 and 2 leading to repeated roots, which are covered in Case 1. Therefore, only the two roots from the quadratic $r^2 + r - 1 = 0$ give us a distinct quadratic equation here.

So, in Subcase 2b, we have one quadratic equation.

Now, are there other possibilities? Let me think.

Possibility 3: One root satisfies $c^2 - 2 = c$ (i.e., is 2 or -1), and the other root satisfies $c^2 - 2$ equals the other root. Wait, but that might be covered in Subcase 2b or in another case.

Wait, let me think again. Suppose one root is 2 (which satisfies $c^2 - 2 = c$), then the other root must satisfy that $c^2 - 2$ is either 2 or the other root. Let me explore this.

Let me suppose that one root is 2, and the other root is $r$. Then, since 2 is a root, $c^2 - 2 = 2^2 - 2 = 2$ is already a root, so that's okay. Now, for the other root $r$, we need that $r^2 - 2$ is a root. Since the roots are 2 and $r$, $r^2 - 2$ must be either 2 or $r$.

Case 3a: $r^2 - 2 = 2$. Then, $r^2 = 4$, so $r = 2$ or $r = -2$. If $r = 2$, then both roots are 2, which is a repeated root, covered in Case 1. If $r = -2$, then the roots are 2 and -2. Let me check if this satisfies the condition. If c = 2, then $c^2 - 2 = 2$, which is a root. If c = -2, then $c^2 - 2 = 4 - 2 = 2$, which is a root. Therefore, this quadratic equation $x^2 - (2 + (-2))x + (2)(-2) = x^2 - 0x - 4 = x^2 - 4 = 0$ satisfies the condition. Therefore, this is another quadratic equation. So, this is a valid case.

Case 3b: $r^2 - 2 = r$. Then, $r^2 - r - 2 = 0$, which gives roots 2 and -1. Therefore, if one root is 2, the other root is either 2 or -1. If it's 2, then repeated root. If it's -1, then the roots are 2 and -1, which is the quadratic we considered in Subcase 2a. Therefore, this case gives us the quadratic $x^2 - x - 2 = 0$ (from roots 2 and -1) and the repeated root case.

Therefore, in Case 3a, we have another quadratic equation: $x^2 - 4 = 0$.

Similarly, we can consider if one root is -1. Let me check that.

Suppose one root is -1, and the other root is $r$. Then, since -1 is a root, $c^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1$, which is already a root. For the other root $r$, $r^2 - 2$ must be either -1 or $r$.

Case 4a: $r^2 - 2 = -1$. Then, $r^2 = 1$, so $r = 1$ or $r = -1$. If $r = -1$, then repeated root. If $r = 1$, then the roots are -1 and 1. Let me check if this satisfies the condition. If c = -1, then $c^2 - 2 = -1$, which is a root. If c = 1, then $c^2 - 2 = 1 - 2 = -1$, which is a root. Therefore, this quadratic equation $x^2 - ( -1 + 1 )x + (-1)(1) = x^2 - 0x -1 = x^2 - 1 = 0$ satisfies the condition. Therefore, this is another quadratic equation: $x^2 - 1 = 0$.

Case 4b: $r^2 - 2 = r$. Then, same as before, leading to roots 2 and -1. Therefore, if one root is -1, the other root is 2 or -1. If it's 2, then the quadratic is $x^2 - x - 2 = 0$, which we already considered. If it's -1, then repeated root.

Therefore, in Case 4a, we have another quadratic equation: $x^2 - 1 = 0$.

So, so far, we have:

From Case 1: two quadratics: $x^2 - 4x + 4$ and $x^2 + 2x + 1$.

From Subcase 2a: one quadratic: $x^2 - x - 2$.

From Subcase 2b: one quadratic: $x^2 + x - 1$.

From Case 3a: one quadratic: $x^2 - 4$.

From Case 4a: one quadratic: $x^2 - 1$.

Wait, but hold on, need to check if these quadratics are distinct. Let me list them:

1. $x^2 - 4x + 4$ (repeated root 2)
2. $x^2 + 2x + 1$ (repeated root -1)
3. $x^2 - x - 2$ (roots 2 and -1)
4. $x^2 + x - 1$ (roots $[-1 \pm \sqrt{5}]/2$)
5. $x^2 - 4$ (roots 2 and -2)
6. $x^2 - 1$ (roots 1 and -1)

So, that's six quadratics. Wait, but need to check if there are overlaps or duplicates. For example, is $x^2 - 4$ different from others? Yes. Similarly, $x^2 - 1$ is different. So, total of six?

Wait, but let me check if there are more possibilities. For example, in Case 3a, we considered one root being 2 and the other being -2. Similarly, in Case 4a, one root being -1 and the other being 1. Are there other possibilities?

Wait, suppose we have a quadratic with roots 1 and -1. Then, as we saw, it satisfies the condition. Similarly, roots 2 and -2. Are there other roots where $c^2 - 2$ is another root?

Wait, let me think. Suppose we have a quadratic with roots 1 and something else. Wait, but we already considered that in Case 4a. Similarly for roots -2 and 2. So, maybe those are the only ones.

But let me check if there are other quadratics where one root is 2 or -1 and the other is something else. For example, suppose we have a quadratic with root 2 and another root r, where r^2 - 2 is either 2 or r. We considered r = -2 and r = -1. But are there other possibilities?

If r^2 - 2 = r, then r = 2 or -1. So, only those. If r^2 - 2 = 2, then r = ±2. So, only those. Therefore, no other possibilities. Similarly for root -1.

Therefore, the quadratics we have are:

1. $x^2 - 4x + 4$
2. $x^2 + 2x + 1$
3. $x^2 - x - 2$
4. $x^2 + x - 1$
5. $x^2 - 4$
6. $x^2 - 1$

So, six quadratics. Wait, but let me check if these are all distinct. Let me check their coefficients:

1. $x^2 - 4x + 4$: a = -4, b = 4
2. $x^2 + 2x + 1$: a = 2, b = 1
3. $x^2 - x - 2$: a = -1, b = -2
4. $x^2 + x - 1$: a = 1, b = -1
5. $x^2 - 4$: a = 0, b = -4
6. $x^2 - 1$: a = 0, b = -1

Yes, all different. So, six quadratics. But wait, let me check if there are more possibilities. For example, could there be quadratics with roots that are not covered in these cases?

Wait, let me think again. The quartic equation we solved earlier had roots -1, 2, and the two roots from $r^2 + r - 1 = 0$. But we considered the case where the roots are those two roots, leading to quadratic $x^2 + x - 1$. Also, we considered quadratics with roots 2 and -2, 1 and -1, 2 and -1, etc. Are there any other quadratics?

Wait, let me think about another possibility. Suppose that the quadratic has roots such that applying the transformation $c \rightarrow c^2 - 2$ cycles between roots. For example, if we have two roots $r_1$ and $r_2$ such that $r_1^2 - 2 = r_2$ and $r_2^2 - 2 = r_1$. Which is exactly the case we considered in Subcase 2b, leading to the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, could there be a cycle of more than two roots? But since the quadratic only has two roots, the transformation can't cycle more than two roots. So, either it's a fixed point (i.e., $c^2 - 2 = c$) or it's a 2-cycle (i.e., $c_1^2 - 2 = c_2$ and $c_2^2 - 2 = c_1$). So, these are the only possibilities. Therefore, we have considered all cases.

But let me check if there are quadratics where both roots are in a 2-cycle. Wait, but we already considered that in Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's covered.

Therefore, total of six quadratics. Wait, but let me check if there are any other quadratics where the roots are not covered in these cases. For example, suppose there is a quadratic with roots that are not 2, -1, 1, -2, or the roots of $x^2 + x - 1$. Is that possible?

Suppose there is a quadratic with roots $r$ and $s$, where neither $r$ nor $s$ satisfy $c^2 - 2 = c$, but $r^2 - 2 = s$ and $s^2 - 2 = r$. But that is exactly the case we considered in Subcase 2b, which gives us the quadratic $x^2 + x - 1$. Therefore, that's already included.

Alternatively, if one root satisfies $c^2 - 2 = c$ (i.e., is 2 or -1) and the other root satisfies $c^2 - 2$ equals the other root. Wait, but that would be covered in Cases 3a and 4a, leading to quadratics with roots 2 and -2, or -1 and 1. So, those are already included.

Therefore, I think we have considered all possibilities. Therefore, total of six quadratics. Wait, but let me check again:

Case 1: 2 quadratics (repeated roots 2 and -1)

Case 2a: 1 quadratic (roots 2 and -1)

Case 2b: 1 quadratic (roots from $x^2 + x - 1$)

Case 3a: 1 quadratic (roots 2 and -2)

Case 4a: 1 quadratic (roots 1 and -1)

Wait, but hold on, Case 2a is the quadratic with roots 2 and -1, which is different from Case 1 quadratics (which are repeated roots). So, total of 2 + 1 + 1 + 1 + 1 = 6 quadratics. But let me check if the quadratic with roots 2 and -1 is unique. Yes, because it's different from the repeated root quadratics.

But wait, in Case 2a, we considered both roots satisfying $c^2 - 2 = c$, which gives roots 2 and -1. So that quadratic is $x^2 - x - 2$. Then, in Case 3a, we have roots 2 and -2, which is another quadratic. Similarly, Case 4a gives roots 1 and -1. So, these are all different.

Therefore, total of six quadratics. However, let me check if there are any duplicates or if I have overcounted.

For example, is the quadratic $x^2 - x - 2$ different from others? Yes. Similarly, $x^2 + x - 1$ is different. The quadratics with roots 2 and -2, 1 and -1, etc., are all different. So, total of six.

But wait, let me check if there are more quadratics. For example, suppose we have a quadratic with roots 1 and something else. Wait, we considered roots 1 and -1. What if there is another root?

Wait, suppose we have a quadratic with root 1 and another root r. Then, since 1 is a root, $1^2 - 2 = -1$ must be a root. Therefore, the other root must be -1. Therefore, the quadratic is $x^2 - (1 + (-1))x + (1)(-1) = x^2 - 0x -1 = x^2 - 1$. Which is already considered in Case 4a. Similarly, if we have a quadratic with root -2, then $(-2)^2 - 2 = 4 - 2 = 2$ must be a root, so the quadratic is $x^2 - ( -2 + 2 )x + (-2)(2) = x^2 - 0x -4 = x^2 - 4$, which is already considered in Case 3a.

Therefore, no other quadratics here.

Therefore, total of six quadratics. Wait, but let me check if there are quadratics with roots that are not covered. For example, suppose there is a quadratic with roots that are not 2, -1, 1, -2, or the roots of $x^2 + x - 1$. Is that possible?

Suppose there is a quadratic with roots $r$ and $s$, where neither $r$ nor $s$ is 2, -1, 1, -2, or the roots of $x^2 + x - 1$. But according to our earlier analysis, the only possibilities are:

Either:

- Both roots are fixed points (i.e., satisfy $c^2 - 2 = c$), which are 2 and -1.

- Or, they form a 2-cycle, which gives the roots from $x^2 + x - 1$.

- Or, one root is a fixed point and the other is such that $c^2 - 2$ is the other root, leading to quadratics with roots 2 and -2, or -1 and 1.

Therefore, there are no other possibilities. Therefore, total of six quadratics.

But wait, let me check if the quadratic $x^2 + x - 1$ is different from others. Yes, because its roots are different from the others. Similarly, the quadratic $x^2 - 1$ has roots 1 and -1, which are different from others.

Wait, but let me check if there are quadratics with roots that are not covered. For example, suppose we have a quadratic with roots $r$ and $r^2 - 2$, but then $r^2 - 2$ must also satisfy that its square minus 2 is another root. Wait, but since there are only two roots, $r$ and $r^2 - 2$, then applying the transformation to $r^2 - 2$ must give either $r$ or $r^2 - 2$. If it gives $r$, then we have the 2-cycle case. If it gives $r^2 - 2$, then it's a fixed point. Therefore, this is covered in the previous cases.

Therefore, I think we have considered all possibilities. Therefore, total of six quadratics. However, let me check again the list:

1. $x^2 - 4x + 4$ (repeated root 2)
2. $x^2 + 2x + 1$ (repeated root -1)
3. $x^2 - x - 2$ (roots 2 and -1)
4. $x^2 + x - 1$ (roots $[-1 \pm \sqrt{5}]/2$)
5. $x^2 - 4$ (roots 2 and -2)
6. $x^2 - 1$ (roots 1 and -1)

Yes, six quadratics. However, let me check if there are any other quadratics where the roots are not covered. For example, suppose we have a quadratic with roots 1 and something else. Wait, we already considered that. Similarly for other roots.

Alternatively, let me check if the quadratic $x^2 + x - 1$ is valid. Let me verify the condition. Let me take the roots $r_1 = [-1 + √5]/2$ and $r_2 = [-1 - √5]/2$. Then, for $r_1$, $r_1^2 - 2 = r_2$, and for $r_2$, $r_2^2 - 2 = r_1$. Therefore, yes, the condition is satisfied. Therefore, this quadratic is valid.

Similarly, for $x^2 - 1$, roots 1 and -1. For 1: $1^2 - 2 = -1$, which is a root. For -1: $(-1)^2 - 2 = -1$, which is a root. Valid.

For $x^2 - 4$, roots 2 and -2. For 2: $2^2 - 2 = 2$, which is a root. For -2: $(-2)^2 - 2 = 2$, which is a root. Valid.

For $x^2 - x - 2$, roots 2 and -1. For 2: $2^2 - 2 = 2$, which is a root. For -1: $(-1)^2 - 2 = -1$, which is a root. Valid.

For the repeated roots: $x^2 - 4x + 4$ (root 2): $2^2 - 2 = 2$, which is the root. Similarly for $x^2 + 2x + 1$ (root -1): $(-1)^2 - 2 = -1$, which is the root. Valid.

Therefore, all six quadratics satisfy the condition. Therefore, the answer should be 6. However, let me check again if there are any duplicates or if I have missed something.

Wait, but hold on. Let me check if the quadratic $x^2 - x - 2$ is different from the others. Yes, because its coefficients are different. Similarly, all others are different. Therefore, total of six quadratics.

But wait, let me check if there are quadratics with different coefficients but same roots. For example, if there is another quadratic with the same roots but different a and b? No, because for a quadratic equation with given roots, the coefficients are uniquely determined by Vieta's formula. Therefore, each set of roots corresponds to exactly one quadratic equation. Therefore, since we have six different sets of roots, there are six different quadratics.

But let me check again the roots:

1. Repeated root 2: {2, 2}
2. Repeated root -1: {-1, -1}
3. Roots 2 and -1: {2, -1}
4. Roots $[-1 \pm \sqrt{5}]/2$: two distinct roots
5. Roots 2 and -2: {2, -2}
6. Roots 1 and -1: {1, -1}

Yes, all different sets of roots. Therefore, six quadratics. Therefore, the answer is 6. However, let me check again if there are more possibilities.

Wait, another thought: suppose that the quadratic has roots such that applying the transformation $c \rightarrow c^2 - 2$ leads to the same root. For example, if a root is 2, then it maps to itself. Similarly for -1. But we already considered those. Also, the roots from $x^2 + x - 1$ form a 2-cycle. So, no other possibilities.

Alternatively, could there be quadratics where one root maps to itself and the other maps to the first root? For example, suppose one root is 2 (fixed), and the other root maps to 2. Then, the other root must satisfy $r^2 - 2 = 2$, so $r^2 = 4$, so r = 2 or -2. If r = 2, then repeated root. If r = -2, then roots 2 and -2, which is already considered. Similarly for -1. Therefore, no new quadratics here.

Therefore, I think the total number is indeed six. However, let me check if the problem says "quadratic equations of the form $x^2 + ax + b = 0$", so different a and b. So, each of these six quadratics have different a and b. Therefore, the answer is 6.

But wait, hold on. Wait, let me check if there are quadratics with the same a and b but different roots? No, because for a quadratic equation, the roots are determined by a and b. So, different roots mean different a and b. Therefore, six different quadratics.

But let me check again the list:

1. $x^2 - 4x + 4$: a = -4, b = 4
2. $x^2 + 2x + 1$: a = 2, b = 1
3. $x^2 - x - 2$: a = -1, b = -2
4. $x^2 + x - 1$: a = 1, b = -1
5. $x^2 - 4$: a = 0, b = -4
6. $x^2 - 1$: a = 0, b = -1

Yes, all different. Therefore, the answer is 6. However, I need to check if there are any other quadratics that I might have missed.

Wait, another thought: suppose that the quadratic has a root which is not in the above list, but when you apply the transformation, it cycles between roots. But since there are only two roots, the only possibilities are fixed points or 2-cycles. Which we have considered. Therefore, no other quadratics.

Alternatively, could there be quadratics with complex roots? The problem didn't specify that the roots have to be real. Wait, but the problem says "whenever c is a root of the equation, c² - 2 is also a root". If the roots are complex, then c² - 2 would also have to be a root. However, the problem didn't specify that the coefficients a and b are real numbers. Wait, but usually, unless specified, we assume that the coefficients are real numbers. Wait, but the problem didn't specify. Wait, but in the original problem statement, it's just asking for quadratic equations of the form $x^2 + ax + b = 0$, so a and b are presumably real numbers, since otherwise, the problem would have mentioned complex numbers. Therefore, we can assume that a and b are real numbers, so the roots are either both real or complex conjugates.

But if the roots are complex, then applying the transformation $c \rightarrow c^2 - 2$ would need to result in another root. However, since complex roots come in conjugate pairs, let me check if such quadratics exist.

Suppose that the quadratic has complex roots. Let me assume that the roots are complex conjugates, say $r$ and $\overline{r}$. Then, if $r$ is a root, then $r^2 - 2$ must be a root. Since the roots are $r$ and $\overline{r}$, then $r^2 - 2$ must be either $r$ or $\overline{r}$.

Case: $r^2 - 2 = r$. Then, $r^2 - r - 2 = 0$. The solutions are real numbers 2 and -1, which are already considered. Therefore, if the roots are complex, this equation would have real roots, which contradicts the assumption of complex roots. Therefore, there are no complex roots in this case.

Alternatively, if $r^2 - 2 = \overline{r}$. Then, since $r$ and $\overline{r}$ are roots, this would imply that $r^2 - 2 = \overline{r}$. Taking conjugate of both sides, we get $\overline{r}^2 - 2 = r$. Therefore, we have the system:

$r^2 - 2 = \overline{r}$

$\overline{r}^2 - 2 = r$

Let me denote $r = p + qi$, where p and q are real numbers. Then, $\overline{r} = p - qi$. Let me substitute into the equations.

First equation: $(p + qi)^2 - 2 = p - qi$

Compute $(p + qi)^2 = p^2 - q^2 + 2p qi$. Therefore:

$p^2 - q^2 + 2p qi - 2 = p - qi$

Equate real and imaginary parts:

Real: $p^2 - q^2 - 2 = p$

Imaginary: $2p q = -q$

Similarly, second equation: $(p - qi)^2 - 2 = p + qi$

Compute $(p - qi)^2 = p^2 - q^2 - 2p qi$. Therefore:

$p^2 - q^2 - 2p qi - 2 = p + qi$

Equate real and imaginary parts:

Real: $p^2 - q^2 - 2 = p$

Imaginary: $-2p q = q$

So, from the imaginary parts:

From first equation: $2p q = -q$. If q ≠ 0, then 2p = -1 ⇒ p = -1/2.

From second equation: -2p q = q. If q ≠ 0, then -2p = 1 ⇒ p = -1/2. So, consistent.

Therefore, p = -1/2. Then, from the real part equation:

$p^2 - q^2 - 2 = p$

Substitute p = -1/2:

$(-1/2)^2 - q^2 - 2 = -1/2$

Compute:

1/4 - q^2 - 2 = -1/2 ⇒ -q^2 - 7/4 = -1/2 ⇒ -q^2 = -1/2 + 7/4 = (-2/4 + 7/4) = 5/4 ⇒ q^2 = -5/4.

But q^2 cannot be negative. Therefore, no solution in this case. Therefore, there are no complex roots satisfying the condition. Therefore, all quadratics must have real roots. Therefore, we don't need to consider complex roots.

Therefore, our previous count of six quadratics is correct.

But let me check once more if there are any other quadratics. For example, suppose we have a quadratic with roots 2 and something else. Wait, we considered that. Similarly for other roots. Therefore, I think six is the correct answer.

But wait, hold on. Let me check if the quadratic $x^2 + x - 1$ is the only one with roots from the quadratic equation $r^2 + r - 1 = 0$. Yes, because that quadratic has two distinct roots, and we already considered that. Therefore, that's one quadratic.

Therefore, total of six quadratics. Therefore, the answer is 6. However, I need to check if the problem counts quadratics with different a and b as different even if they have the same roots. But since the problem says "quadratic equations of the form $x^2 + ax + b = 0$", each different a and b gives a different equation, even if they have the same roots. Wait, no, if two quadratics have the same roots, they are the same equation up to multiplication by a constant. But since the leading coefficient is 1, they are the same equation. Therefore, each set of roots corresponds to exactly one quadratic equation. Therefore, since we have six different sets of roots, there are six different equations.

Therefore, the answer is 6. However, I need to check again if I have considered all possibilities. Let me think again.

Wait, another way to approach this problem is to consider that the set of roots must be closed under the transformation $c \rightarrow c^2 - 2$. Since the quadratic has two roots, the set {r1, r2} must be closed under this transformation. Therefore, the possible cases are:

1. {r} where r is a fixed point (i.e., r^2 - 2 = r). These are the roots 2 and -1. So, quadratics with repeated roots 2 or -1.

2. {r, s} where r^2 - 2 = s and s^2 - 2 = r. This is the 2-cycle case, leading to the quadratic $x^2 + x - 1$.

3. {r, s} where one of them is a fixed point and the other is such that applying the transformation gives the fixed point. For example, {2, -2} because -2^2 - 2 = 2, and 2^2 - 2 = 2. Similarly, {-1, 1} because 1^2 - 2 = -1, and (-1)^2 - 2 = -1. Therefore, these are quadratics with roots {2, -2} and {-1, 1}.

Therefore, these are the four cases:

- Two fixed points (repeated roots): 2 and -1.

- One 2-cycle: {r, s} from $x^2 + x - 1$.

- Two cases where one root is a fixed point and the other maps to it: {2, -2} and {-1, 1}.

But wait, the case where one root is a fixed point and the other maps to it gives two quadratics: {2, -2} and {-1, 1}. However, the case where both roots are fixed points gives two quadratics (repeated roots). And the case of a 2-cycle gives one quadratic. Therefore, total of 2 + 1 + 2 = 5? Wait, but earlier I had six. Where is the discrepancy?

Wait, in the previous analysis, I considered:

Case 1: two quadratics (repeated roots 2 and -1)

Case 2a: one quadratic (roots 2 and -1)

Case 2b: one quadratic (2-cycle)

Case 3a: one quadratic (roots 2 and -2)

Case 4a: one quadratic (roots -1 and 1)

So, total of 2 + 1 + 1 + 1 + 1 = 6. But according to this new approach, it's 2 (repeated roots) + 1 (2-cycle) + 2 (fixed point and mapped root) = 5. Where is the difference?

Ah, because in the previous analysis, the quadratic with roots 2 and -1 is considered in both Case 1 (repeated roots) and Case 2a (distinct roots). But actually, the quadratic with roots 2 and -1 is different from the repeated roots quadratics. Therefore, in the new approach, the case where both roots are fixed points (i.e., {2, -1}) is a distinct case from the repeated roots. Therefore, the total cases are:

- Repeated roots: 2 quadratics.

- Distinct roots with both fixed points: 1 quadratic.

- 2-cycle: 1 quadratic.

- Distinct roots with one fixed point and one mapped root: 2 quadratics.

Total: 2 + 1 + 1 + 2 = 6. Yes, that matches. Therefore, the answer is 6.

Therefore, I think the correct answer is 6. However, to be thorough, let me check once more.

List of quadratics:

1. $x^2 - 4x + 4$ (repeated root 2)

2. $x^2 + 2x + 1$ (repeated root -1)

3. $x^2 - x - 2$ (roots 2 and -1)

4. $x^2 + x - 1$ (2-cycle)

5. $x^2 - 4$ (roots 2 and -2)

6. $x^2 - 1$ (roots 1 and -1)

Yes, six quadratics. Therefore, the answer is 6.

But wait, let me check if the quadratic $x^2 - x - 2$ is different from the others. Yes, because its roots are 2 and -1, which are different from the other quadratics. Therefore, all six are unique.

Therefore, I think the answer is 6. However, I recall that sometimes problems like this might have fewer solutions due to some constraints, but in this case, I think all possibilities are covered.

Therefore, my final answer is 6, so the number of such quadratic equations is \boxed{6}.

**Final Answer**
\boxed{6}

To determine the number of quadratic equations of the form $ x^2 + ax + b = 0 $ such that **whenever $ c $ is a root, $ c^2 - 2 $ is also a root**, we analyze the implications of this condition on the roots of the equation.

---

### Step 1: Understanding the Condition

Let the roots of the quadratic be $ r_1 $ and $ r_2 $. The condition implies that for each root $ c $, the value $ c^2 - 2 $ must also be a root. Since a quadratic has only two roots, this condition must be satisfied for both $ r_1 $ and $ r_2 $.

This leads to the following possibilities:

- **Case 1:** The quadratic has a **repeated root** $ r $, i.e., $ r_1 = r_2 = r $. Then, the condition becomes $ r^2 - 2 = r $, which simplifies to:
  $$
  r^2 - r - 2 = 0
  $$
  Solving this quadratic gives:
  $$
  r = 2 \quad \text{or} \quad r = -1
  $$
  So, the possible repeated root quadratics are:
  $$
  (x - 2)^2 = x^2 - 4x + 4 \quad \text{and} \quad (x + 1)^2 = x^2 + 2x + 1
  $$

- **Case 2:** The quadratic has **distinct roots** $ r_1 $ and $ r_2 $. Then, the condition implies:
  $$
  r_1^2 - 2 = r_2 \quad \text{and} \quad r_2^2 - 2 = r_1
  $$
  This system of equations leads to a quartic equation:
  $$
  r^4 - 4r^2 - r + 2 = 0
  $$
  Factoring this gives:
  $$
  (r + 1)(r - 2)(r^2 + r - 1) = 0
  $$
  The roots are:
  $$
  r = -1, \quad r = 2, \quad r = \frac{-1 \pm \sqrt{5}}{2}
  $$
  The roots $ \frac{-1 \pm \sqrt{5}}{2} $ form a **2-cycle** under the transformation $ c \mapsto c^2 - 2 $, i.e., $ r_1^2 - 2 = r_2 $ and $ r_2^2 - 2 = r_1 $. The corresponding quadratic is:
  $$
  x^2 + x - 1
  $$

- **Case 3:** One root is a fixed point (i.e., $ c^2 - 2 = c $), and the other root maps to the fixed point under the transformation. This leads to the following possibilities:

  - If one root is $ 2 $, then the other root must satisfy $ r^2 - 2 = 2 $, so $ r^2 = 4 $, giving $ r = \pm 2 $. The quadratic with roots $ 2 $ and $ -2 $ is:
    $$
    x^2 - 4
    $$

  - If one root is $ -1 $, then the other root must satisfy $ r^2 - 2 = -1 $, so $ r^2 = 1 $, giving $ r = \pm 1 $. The quadratic with roots $ -1 $ and $ 1 $ is:
    $$
    x^2 - 1
    $$

---

### Step 2: Counting the Distinct Quadratics

We now list all the distinct quadratic equations that satisfy the condition:

1. $ x^2 - 4x + 4 $ (repeated root 2)
2. $ x^2 + 2x + 1 $ (repeated root -1)
3. $ x^2 - x - 2 $ (roots 2 and -1)
4. $ x^2 + x - 1 $ (roots $ \frac{-1 \pm \sqrt{5}}{2} $)
5. $ x^2 - 4 $ (roots 2 and -2)
6. $ x^2 - 1 $ (roots 1 and -1)

Each of these quadratics has **distinct coefficients** $ a $ and $ b $, and thus represents a **unique quadratic equation** of the form $ x^2 + ax + b = 0 $.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the number of quadratic equations of the form $x^2 + ax + b = 0$ such that whenever $c$ is a root, $c^2 - 2$ is also a root. Hmm, let me think about how to approach this.

First, let me recall that a quadratic equation has two roots (counting multiplicity). Let me denote the roots as $r_1$ and $r_2$. According to Vieta's formulas, we have:

$r_1 + r_2 = -a$ and $r_1 r_2 = b$.

Now, the condition given is that if $c$ is a root, then $c^2 - 2$ is also a root. So, for each root $c$, $c^2 - 2$ must be another root. Since there are only two roots, this condition must hold for both roots. Let me consider the possibilities.

Case 1: The quadratic has a repeated root. That is, $r_1 = r_2$. Then, if $c = r_1$, then $c^2 - 2$ must also be a root. But since there's only one root, $c^2 - 2$ must equal $c$. So, solving $c^2 - 2 = c$, which gives $c^2 - c - 2 = 0$. Solving this quadratic equation:

$c = [1 \pm \sqrt{1 + 8}]/2 = [1 \pm 3]/2$, so $c = 2$ or $c = -1$.

Therefore, if the quadratic has a repeated root, it must be either 2 or -1. Let me check if these satisfy the condition.

If the root is 2, then $c^2 - 2 = 4 - 2 = 2$, which is the same root. So that works. Similarly, if the root is -1, then $c^2 - 2 = 1 - 2 = -1$, which is also the same root. So both 2 and -1 are valid repeated roots. Therefore, there are two quadratics in this case: $(x - 2)^2 = x^2 - 4x + 4$ and $(x + 1)^2 = x^2 + 2x + 1$. So that's two equations.

Case 2: The quadratic has two distinct roots, $r_1$ and $r_2$. Then, the condition implies that if $r_1$ is a root, then $r_1^2 - 2$ must be a root. Similarly, if $r_2$ is a root, then $r_2^2 - 2$ must be a root. Since there are only two roots, this means that $r_1^2 - 2$ must be either $r_1$ or $r_2$, and similarly for $r_2^2 - 2$.

So, let me consider the possibilities for each root:

For root $r_1$:

Either $r_1^2 - 2 = r_1$ or $r_1^2 - 2 = r_2$.

Similarly for root $r_2$:

Either $r_2^2 - 2 = r_2$ or $r_2^2 - 2 = r_1$.

But since there are two roots, we need to consider combinations of these possibilities. Let me think about possible scenarios.

Subcase 2a: Both roots satisfy $c^2 - 2 = c$. Then, both roots are solutions to $c^2 - c - 2 = 0$, which we already solved earlier, giving roots 2 and -1. But if both roots are 2 and -1, then the quadratic would be $(x - 2)(x + 1) = x^2 - x - 2$. However, this is a quadratic with distinct roots. Wait, but in this case, the roots are 2 and -1. Let me check if this satisfies the condition. If c is 2, then $c^2 - 2 = 4 - 2 = 2$, which is already a root. If c is -1, then $c^2 - 2 = 1 - 2 = -1$, which is also a root. So this quadratic does satisfy the condition. However, this quadratic was not considered in Case 1 because it has distinct roots. Wait, but in Case 1, we considered repeated roots. So this quadratic is a separate case. Therefore, this is another quadratic. But wait, but in this case, both roots satisfy $c^2 - 2 = c$, so this is a valid quadratic. However, this quadratic is different from the ones in Case 1. So, this is another quadratic. Wait, but hold on, if both roots satisfy $c^2 - 2 = c$, then the quadratic equation would have roots 2 and -1. So that's one quadratic. But is there another quadratic in this subcase? No, because the equation $c^2 - c - 2 = 0$ has exactly two roots, so the quadratic with roots 2 and -1 is unique. Therefore, this is one quadratic in Subcase 2a.

Subcase 2b: One root satisfies $c^2 - 2 = c$, and the other root satisfies $c^2 - 2$ equals the other root. Wait, let me think again. Let me consider that for each root, $c^2 - 2$ must be another root. So, if one root is 2 (which satisfies $c^2 - 2 = c$), then the other root must satisfy that if you take $c^2 - 2$, it must be either 2 or the other root. Similarly for the other root.

Alternatively, maybe there's a better way to approach this. Let me think.

Suppose that the roots are $r_1$ and $r_2$. Then, the condition implies that:

If $r_1$ is a root, then $r_1^2 - 2$ is also a root. Therefore, $r_1^2 - 2$ must be either $r_1$ or $r_2$. Similarly, $r_2^2 - 2$ must be either $r_1$ or $r_2$.

So, there are several possibilities here. Let me consider different possibilities:

Possibility 1: $r_1^2 - 2 = r_1$ and $r_2^2 - 2 = r_2$. Then, both roots satisfy $c^2 - c - 2 = 0$, so they are 2 and -1. This is the quadratic we considered in Subcase 2a.

Possibility 2: $r_1^2 - 2 = r_2$ and $r_2^2 - 2 = r_1$. So, this is a system of equations. Let me write that down:

$r_1^2 - 2 = r_2$ ...(1)

$r_2^2 - 2 = r_1$ ...(2)

So, substituting equation (1) into equation (2):

From (1): $r_2 = r_1^2 - 2$. Substitute into (2):

$(r_1^2 - 2)^2 - 2 = r_1$

Expanding that:

$r_1^4 - 4r_1^2 + 4 - 2 = r_1$

Simplify:

$r_1^4 - 4r_1^2 + 2 = r_1$

Bring all terms to one side:

$r_1^4 - 4r_1^2 - r_1 + 2 = 0$

Hmm, quartic equation. Let me try to factor this. Maybe there are rational roots. Let me apply Rational Root Theorem. Possible rational roots are ±1, ±2.

Testing r = 1: 1 - 4 - 1 + 2 = -2 ≠ 0

Testing r = -1: 1 - 4 + 1 + 2 = 0. Oh, r = -1 is a root.

So, we can factor (r + 1) out of the quartic.

Let me perform polynomial division or use synthetic division.

Dividing $r^4 - 4r^2 - r + 2$ by (r + 1):

Using synthetic division for root -1:

Coefficients: 1 (r^4), 0 (r^3), -4 (r^2), -1 (r), 2 (constant)

Bring down the 1.

Multiply by -1: -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: 1. Add to next coefficient: -4 + 1 = -3.

Multiply by -1: 3. Add to next coefficient: -1 + 3 = 2.

Multiply by -1: -2. Add to last coefficient: 2 + (-2) = 0. Perfect.

So, the quartic factors as (r + 1)(r^3 - r^2 - 3r + 2).

Now, let's factor the cubic: $r^3 - r^2 - 3r + 2$.

Again, try rational roots: ±1, ±2.

Testing r = 1: 1 - 1 - 3 + 2 = -1 ≠ 0.

Testing r = 2: 8 - 4 - 6 + 2 = 0. Yes, r = 2 is a root.

Divide the cubic by (r - 2):

Using synthetic division:

Coefficients: 1 (r^3), -1 (r^2), -3 (r), 2 (constant)

Root at 2:

Bring down 1.

Multiply by 2: 2. Add to next coefficient: -1 + 2 = 1.

Multiply by 2: 2. Add to next coefficient: -3 + 2 = -1.

Multiply by 2: -2. Add to last coefficient: 2 + (-2) = 0.

So, cubic factors as (r - 2)(r^2 + r - 1).

Therefore, the quartic factors as (r + 1)(r - 2)(r^2 + r - 1).

Therefore, the roots of the quartic equation are r = -1, r = 2, and roots of $r^2 + r - 1 = 0$, which are $r = [-1 \pm \sqrt{1 + 4}]/2 = [-1 \pm \sqrt{5}]/2$.

So, the solutions to the quartic equation are r = -1, 2, and $[-1 \pm \sqrt{5}]/2$.

Now, going back to our system of equations (1) and (2):

We have that $r_1$ and $r_2$ satisfy $r_1^2 - 2 = r_2$ and $r_2^2 - 2 = r_1$. So, the solutions for $r_1$ and $r_2$ can be:

Either:

- $r_1 = -1$, then from equation (1): $r_2 = (-1)^2 - 2 = 1 - 2 = -1$. So, both roots are -1. But this is a repeated root, which was considered in Case 1. However, in this case, since we are in Subcase 2b where we are considering distinct roots? Wait, no. Wait, in Subcase 2b, we are considering the case where one root satisfies $c^2 - 2 = c$ and the other satisfies $c^2 - 2$ equals the other root. Wait, but if both roots are -1, then they are the same root, so that's a repeated root. But in this case, we are in Case 2 where we have two distinct roots. Therefore, this solution (both roots -1) is actually covered in Case 1, so we need to exclude it here.

Similarly, if $r_1 = 2$, then from equation (1): $r_2 = 2^2 - 2 = 4 - 2 = 2$. Again, same root, so repeated root. So again, covered in Case 1.

Therefore, the other roots are $r = [-1 \pm \sqrt{5}]/2$. Let me denote them as $r_1 = [-1 + \sqrt{5}]/2$ and $r_2 = [-1 - \sqrt{5}]/2$. Let me check if they satisfy the equations.

First, check if $r_1^2 - 2 = r_2$:

Compute $r_1^2$:

Let me compute $r_1 = (-1 + √5)/2$. Then,

$r_1^2 = [(-1)^2 + 2*(-1)*(√5) + (√5)^2]/4 = [1 - 2√5 + 5]/4 = (6 - 2√5)/4 = (3 - √5)/2.

Then, $r_1^2 - 2 = (3 - √5)/2 - 2 = (3 - √5 - 4)/2 = (-1 - √5)/2 = r_2$. Perfect, that's correct.

Similarly, check $r_2^2 - 2 = r_1$:

Compute $r_2 = (-1 - √5)/2$. Then,

$r_2^2 = [(-1)^2 + 2*(-1)*(-√5) + (√5)^2]/4 = [1 + 2√5 + 5]/4 = (6 + 2√5)/4 = (3 + √5)/2.

Then, $r_2^2 - 2 = (3 + √5)/2 - 2 = (3 + √5 - 4)/2 = (-1 + √5)/2 = r_1$. Correct.

Therefore, these two roots satisfy the condition. Therefore, the quadratic equation with roots $r_1$ and $r_2$ is:

$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$.

Compute $r_1 + r_2$:

$[(-1 + √5)/2 + (-1 - √5)/2] = (-2)/2 = -1$.

Compute $r_1 r_2$:

Multiply the two roots:

$[(-1 + √5)/2] * [(-1 - √5)/2] = [(-1)^2 - (√5)^2]/4 = [1 - 5]/4 = (-4)/4 = -1$.

Therefore, the quadratic equation is $x^2 - (-1)x + (-1) = x^2 + x - 1 = 0$. So that's another quadratic equation.

Therefore, in this Subcase 2b, we have one quadratic equation: $x^2 + x - 1 = 0$.

Wait, but hold on, are there other possibilities? For example, if we take $r_1 = [-1 + √5]/2$ and $r_2 = [-1 - √5]/2$, which gives us the equation above. But since the roots are distinct, this is one quadratic. However, could there be another quadratic if we take different roots? For example, if we take $r_1 = [-1 + √5]/2$ and $r_2 = something else? Wait, no, because the system of equations only gives us these roots. So, only one quadratic here.

But wait, let me check if there are other solutions. The quartic equation had roots -1, 2, and the two roots from the quadratic. But we already considered -1 and 2 leading to repeated roots, which are covered in Case 1. Therefore, only the two roots from the quadratic $r^2 + r - 1 = 0$ give us a distinct quadratic equation here.

So, in Subcase 2b, we have one quadratic equation.

Now, are there other possibilities? Let me think.

Possibility 3: One root satisfies $c^2 - 2 = c$ (i.e., is 2 or -1), and the other root satisfies $c^2 - 2$ equals the other root. Wait, but that might be covered in Subcase 2b or in another case.

Wait, let me think again. Suppose one root is 2 (which satisfies $c^2 - 2 = c$), then the other root must satisfy that $c^2 - 2$ is either 2 or the other root. Let me explore this.

Let me suppose that one root is 2, and the other root is $r$. Then, since 2 is a root, $c^2 - 2 = 2^2 - 2 = 2$ is already a root, so that's okay. Now, for the other root $r$, we need that $r^2 - 2$ is a root. Since the roots are 2 and $r$, $r^2 - 2$ must be either 2 or $r$.

Case 3a: $r^2 - 2 = 2$. Then, $r^2 = 4$, so $r = 2$ or $r = -2$. If $r = 2$, then both roots are 2, which is a repeated root, covered in Case 1. If $r = -2$, then the roots are 2 and -2. Let me check if this satisfies the condition. If c = 2, then $c^2 - 2 = 2$, which is a root. If c = -2, then $c^2 - 2 = 4 - 2 = 2$, which is a root. Therefore, this quadratic equation $x^2 - (2 + (-2))x + (2)(-2) = x^2 - 0x - 4 = x^2 - 4 = 0$ satisfies the condition. Therefore, this is another quadratic equation. So, this is a valid case.

Case 3b: $r^2 - 2 = r$. Then, $r^2 - r - 2 = 0$, which gives roots 2 and -1. Therefore, if one root is 2, the other root is either 2 or -1. If it's 2, then repeated root. If it's -1, then the roots are 2 and -1, which is the quadratic we considered in Subcase 2a. Therefore, this case gives us the quadratic $x^2 - x - 2 = 0$ (from roots 2 and -1) and the repeated root case.

Therefore, in Case 3a, we have another quadratic equation: $x^2 - 4 = 0$.

Similarly, we can consider if one root is -1. Let me check that.

Suppose one root is -1, and the other root is $r$. Then, since -1 is a root, $c^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1$, which is already a root. For the other root $r$, $r^2 - 2$ must be either -1 or $r$.

Case 4a: $r^2 - 2 = -1$. Then, $r^2 = 1$, so $r = 1$ or $r = -1$. If $r = -1$, then repeated root. If $r = 1$, then the roots are -1 and 1. Let me check if this satisfies the condition. If c = -1, then $c^2 - 2 = -1$, which is a root. If c = 1, then $c^2 - 2 = 1 - 2 = -1$, which is a root. Therefore, this quadratic equation $x^2 - ( -1 + 1 )x + (-1)(1) = x^2 - 0x -1 = x^2 - 1 = 0$ satisfies the condition. Therefore, this is another quadratic equation: $x^2 - 1 = 0$.

Case 4b: $r^2 - 2 = r$. Then, same as before, leading to roots 2 and -1. Therefore, if one root is -1, the other root is 2 or -1. If it's 2, then the quadratic is $x^2 - x - 2 = 0$, which we already considered. If it's -1, then repeated root.

Therefore, in Case 4a, we have another quadratic equation: $x^2 - 1 = 0$.

So, so far, we have:

From Case 1: two quadratics: $x^2 - 4x + 4$ and $x^2 + 2x + 1$.

From Subcase 2a: one quadratic: $x^2 - x - 2$.

From Subcase 2b: one quadratic: $x^2 + x - 1$.

From Case 3a: one quadratic: $x^2 - 4$.

From Case 4a: one quadratic: $x^2 - 1$.

Wait, but hold on, need to check if these quadratics are distinct. Let me list them:

1. $x^2 - 4x + 4$ (repeated root 2)
2. $x^2 + 2x + 1$ (repeated root -1)
3. $x^2 - x - 2$ (roots 2 and -1)
4. $x^2 + x - 1$ (roots $[-1 \pm \sqrt{5}]/2$)
5. $x^2 - 4$ (roots 2 and -2)
6. $x^2 - 1$ (roots 1 and -1)

So, that's six quadratics. Wait, but need to check if there are overlaps or duplicates. For example, is $x^2 - 4$ different from others? Yes. Similarly, $x^2 - 1$ is different. So, total of six?

Wait, but let me check if there are more possibilities. For example, in Case 3a, we considered one root being 2 and the other being -2. Similarly, in Case 4a, one root being -1 and the other being 1. Are there other possibilities?

Wait, suppose we have a quadratic with roots 1 and -1. Then, as we saw, it satisfies the condition. Similarly, roots 2 and -2. Are there other roots where $c^2 - 2$ is another root?

Wait, let me think. Suppose we have a quadratic with roots 1 and something else. Wait, but we already considered that in Case 4a. Similarly for roots -2 and 2. So, maybe those are the only ones.

But let me check if there are other quadratics where one root is 2 or -1 and the other is something else. For example, suppose we have a quadratic with root 2 and another root r, where r^2 - 2 is either 2 or r. We considered r = -2 and r = -1. But are there other possibilities?

If r^2 - 2 = r, then r = 2 or -1. So, only those. If r^2 - 2 = 2, then r = ±2. So, only those. Therefore, no other possibilities. Similarly for root -1.

Therefore, the quadratics we have are:

1. $x^2 - 4x + 4$
2. $x^2 + 2x + 1$
3. $x^2 - x - 2$
4. $x^2 + x - 1$
5. $x^2 - 4$
6. $x^2 - 1$

So, six quadratics. Wait, but let me check if these are all distinct. Let me check their coefficients:

1. $x^2 - 4x + 4$: a = -4, b = 4
2. $x^2 + 2x + 1$: a = 2, b = 1
3. $x^2 - x - 2$: a = -1, b = -2
4. $x^2 + x - 1$: a = 1, b = -1
5. $x^2 - 4$: a = 0, b = -4
6. $x^2 - 1$: a = 0, b = -1

Yes, all different. So, six quadratics. But wait, let me check if there are more possibilities. For example, could there be quadratics with roots that are not covered in these cases?

Wait, let me think again. The quartic equation we solved earlier had roots -1, 2, and the two roots from $r^2 + r - 1 = 0$. But we considered the case where the roots are those two roots, leading to quadratic $x^2 + x - 1$. Also, we considered quadratics with roots 2 and -2, 1 and -1, 2 and -1, etc. Are there any other quadratics?

Wait, let me think about another possibility. Suppose that the quadratic has roots such that applying the transformation $c \rightarrow c^2 - 2$ cycles between roots. For example, if we have two roots $r_1$ and $r_2$ such that $r_1^2 - 2 = r_2$ and $r_2^2 - 2 = r_1$. Which is exactly the case we considered in Subcase 2b, leading to the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, could there be a cycle of more than two roots? But since the quadratic only has two roots, the transformation can't cycle more than two roots. So, either it's a fixed point (i.e., $c^2 - 2 = c$) or it's a 2-cycle (i.e., $c_1^2 - 2 = c_2$ and $c_2^2 - 2 = c_1$). So, these are the only possibilities. Therefore, we have considered all cases.

But let me check if there are quadratics where both roots are in a 2-cycle. Wait, but we already considered that in Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's covered.

Therefore, total of six quadratics. Wait, but let me check if there are any other quadratics where the roots are not covered in these cases. For example, suppose there is a quadratic with roots that are not 2, -1, 1, -2, or the roots of $x^2 + x - 1$. Is that possible?

Suppose there is a quadratic with roots $r$ and $s$, where neither $r$ nor $s$ satisfy $c^2 - 2 = c$, but $r^2 - 2 = s$ and $s^2 - 2 = r$. But that is exactly the case we considered in Subcase 2b, which gives us the quadratic $x^2 + x - 1$. Therefore, that's already included.

Alternatively, if one root satisfies $c^2 - 2 = c$ (i.e., is 2 or -1) and the other root satisfies $c^2 - 2$ equals the other root. Wait, but that would be covered in Cases 3a and 4a, leading to quadratics with roots 2 and -2, or -1 and 1. So, those are already included.

Therefore, I think we have considered all possibilities. Therefore, total of six quadratics. Wait, but let me check again:

Case 1: 2 quadratics (repeated roots 2 and -1)

Case 2a: 1 quadratic (roots 2 and -1)

Case 2b: 1 quadratic (roots from $x^2 + x - 1$)

Case 3a: 1 quadratic (roots 2 and -2)

Case 4a: 1 quadratic (roots 1 and -1)

Wait, but hold on, Case 2a is the quadratic with roots 2 and -1, which is different from Case 1 quadratics (which are repeated roots). So, total of 2 + 1 + 1 + 1 + 1 = 6 quadratics. But let me check if the quadratic with roots 2 and -1 is unique. Yes, because it's different from the repeated root quadratics.

But wait, in Case 2a, we considered both roots satisfying $c^2 - 2 = c$, which gives roots 2 and -1. So that quadratic is $x^2 - x - 2$. Then, in Case 3a, we have roots 2 and -2, which is another quadratic. Similarly, Case 4a gives roots 1 and -1. So, these are all different.

Therefore, total of six quadratics. However, let me check if there are any duplicates or if I have overcounted.

For example, is the quadratic $x^2 - x - 2$ different from others? Yes. Similarly, $x^2 + x - 1$ is different. The quadratics with roots 2 and -2, 1 and -1, etc., are all different. So, total of six.

But wait, let me check if there are more quadratics. For example, suppose we have a quadratic with roots 1 and something else. Wait, we considered roots 1 and -1. What if there is another root?

Wait, suppose we have a quadratic with root 1 and another root r. Then, since 1 is a root, $1^2 - 2 = -1$ must be a root. Therefore, the other root must be -1. Therefore, the quadratic is $x^2 - (1 + (-1))x + (1)(-1) = x^2 - 0x -1 = x^2 - 1$. Which is already considered in Case 4a. Similarly, if we have a quadratic with root -2, then $(-2)^2 - 2 = 4 - 2 = 2$ must be a root, so the quadratic is $x^2 - ( -2 + 2 )x + (-2)(2) = x^2 - 0x -4 = x^2 - 4$, which is already considered in Case 3a.

Therefore, no other quadratics here.

Therefore, total of six quadratics. Wait, but let me check if there are quadratics with roots that are not covered. For example, suppose there is a quadratic with roots that are not 2, -1, 1, -2, or the roots of $x^2 + x - 1$. Is that possible?

Suppose there is a quadratic with roots $r$ and $s$, where neither $r$ nor $s$ is 2, -1, 1, -2, or the roots of $x^2 + x - 1$. But according to our earlier analysis, the only possibilities are:

Either:

- Both roots are fixed points (i.e., satisfy $c^2 - 2 = c$), which are 2 and -1.

- Or, they form a 2-cycle, which gives the roots from $x^2 + x - 1$.

- Or, one root is a fixed point and the other is such that $c^2 - 2$ is the other root, leading to quadratics with roots 2 and -2, or -1 and 1.

Therefore, there are no other possibilities. Therefore, total of six quadratics.

But wait, let me check if the quadratic $x^2 + x - 1$ is different from others. Yes, because its roots are different from the others. Similarly, the quadratic $x^2 - 1$ has roots 1 and -1, which are different from others.

Wait, but let me check if there are quadratics with roots that are not covered. For example, suppose we have a quadratic with roots $r$ and $r^2 - 2$, but then $r^2 - 2$ must also satisfy that its square minus 2 is another root. Wait, but since there are only two roots, $r$ and $r^2 - 2$, then applying the transformation to $r^2 - 2$ must give either $r$ or $r^2 - 2$. If it gives $r$, then we have the 2-cycle case. If it gives $r^2 - 2$, then it's a fixed point. Therefore, this is covered in the previous cases.

Therefore, I think we have considered all possibilities. Therefore, total of six quadratics. However, let me check again the list:

1. $x^2 - 4x + 4$ (repeated root 2)
2. $x^2 + 2x + 1$ (repeated root -1)
3. $x^2 - x - 2$ (roots 2 and -1)
4. $x^2 + x - 1$ (roots $[-1 \pm \sqrt{5}]/2$)
5. $x^2 - 4$ (roots 2 and -2)
6. $x^2 - 1$ (roots 1 and -1)

Yes, six quadratics. However, let me check if there are any other quadratics where the roots are not covered. For example, suppose we have a quadratic with roots 1 and something else. Wait, we already considered that. Similarly for other roots.

Alternatively, let me check if the quadratic $x^2 + x - 1$ is valid. Let me verify the condition. Let me take the roots $r_1 = [-1 + √5]/2$ and $r_2 = [-1 - √5]/2$. Then, for $r_1$, $r_1^2 - 2 = r_2$, and for $r_2$, $r_2^2 - 2 = r_1$. Therefore, yes, the condition is satisfied. Therefore, this quadratic is valid.

Similarly, for $x^2 - 1$, roots 1 and -1. For 1: $1^2 - 2 = -1$, which is a root. For -1: $(-1)^2 - 2 = -1$, which is a root. Valid.

For $x^2 - 4$, roots 2 and -2. For 2: $2^2 - 2 = 2$, which is a root. For -2: $(-2)^2 - 2 = 2$, which is a root. Valid.

For $x^2 - x - 2$, roots 2 and -1. For 2: $2^2 - 2 = 2$, which is a root. For -1: $(-1)^2 - 2 = -1$, which is a root. Valid.

For the repeated roots: $x^2 - 4x + 4$ (root 2): $2^2 - 2 = 2$, which is the root. Similarly for $x^2 + 2x + 1$ (root -1): $(-1)^2 - 2 = -1$, which is the root. Valid.

Therefore, all six quadratics satisfy the condition. Therefore, the answer should be 6. However, let me check again if there are any duplicates or if I have missed something.

Wait, but hold on. Let me check if the quadratic $x^2 - x - 2$ is different from the others. Yes, because its coefficients are different. Similarly, all others are different. Therefore, total of six quadratics.

But wait, let me check if there are quadratics with different coefficients but same roots. For example, if there is another quadratic with the same roots but different a and b? No, because for a quadratic equation with given roots, the coefficients are uniquely determined by Vieta's formula. Therefore, each set of roots corresponds to exactly one quadratic equation. Therefore, since we have six different sets of roots, there are six different quadratics.

But let me check again the roots:

1. Repeated root 2: {2, 2}
2. Repeated root -1: {-1, -1}
3. Roots 2 and -1: {2, -1}
4. Roots $[-1 \pm \sqrt{5}]/2$: two distinct roots
5. Roots 2 and -2: {2, -2}
6. Roots 1 and -1: {1, -1}

Yes, all different sets of roots. Therefore, six quadratics. Therefore, the answer is 6. However, let me check again if there are more possibilities.

Wait, another thought: suppose that the quadratic has roots such that applying the transformation $c \rightarrow c^2 - 2$ leads to the same root. For example, if a root is 2, then it maps to itself. Similarly for -1. But we already considered those. Also, the roots from $x^2 + x - 1$ form a 2-cycle. So, no other possibilities.

Alternatively, could there be quadratics where one root maps to itself and the other maps to the first root? For example, suppose one root is 2 (fixed), and the other root maps to 2. Then, the other root must satisfy $r^2 - 2 = 2$, so $r^2 = 4$, so r = 2 or -2. If r = 2, then repeated root. If r = -2, then roots 2 and -2, which is already considered. Similarly for -1. Therefore, no new quadratics here.

Therefore, I think the total number is indeed six. However, let me check if the problem says "quadratic equations of the form $x^2 + ax + b = 0$", so different a and b. So, each of these six quadratics have different a and b. Therefore, the answer is 6.

But wait, hold on. Wait, let me check if there are quadratics with the same a and b but different roots? No, because for a quadratic equation, the roots are determined by a and b. So, different roots mean different a and b. Therefore, six different quadratics.

But let me check again the list:

1. $x^2 - 4x + 4$: a = -4, b = 4
2. $x^2 + 2x + 1$: a = 2, b = 1
3. $x^2 - x - 2$: a = -1, b = -2
4. $x^2 + x - 1$: a = 1, b = -1
5. $x^2 - 4$: a = 0, b = -4
6. $x^2 - 1$: a = 0, b = -1

Yes, all different. Therefore, the answer is 6. However, I need to check if there are any other quadratics that I might have missed.

Wait, another thought: suppose that the quadratic has a root which is not in the above list, but when you apply the transformation, it cycles between roots. But since there are only two roots, the only possibilities are fixed points or 2-cycles. Which we have considered. Therefore, no other quadratics.

Alternatively, could there be quadratics with complex roots? The problem didn't specify that the roots have to be real. Wait, but the problem says "whenever c is a root of the equation, c² - 2 is also a root". If the roots are complex, then c² - 2 would also have to be a root. However, the problem didn't specify that the coefficients a and b are real numbers. Wait, but usually, unless specified, we assume that the coefficients are real numbers. Wait, but the problem didn't specify. Wait, but in the original problem statement, it's just asking for quadratic equations of the form $x^2 + ax + b = 0$, so a and b are presumably real numbers, since otherwise, the problem would have mentioned complex numbers. Therefore, we can assume that a and b are real numbers, so the roots are either both real or complex conjugates.

But if the roots are complex, then applying the transformation $c \rightarrow c^2 - 2$ would need to result in another root. However, since complex roots come in conjugate pairs, let me check if such quadratics exist.

Suppose that the quadratic has complex roots. Let me assume that the roots are complex conjugates, say $r$ and $\overline{r}$. Then, if $r$ is a root, then $r^2 - 2$ must be a root. Since the roots are $r$ and $\overline{r}$, then $r^2 - 2$ must be either $r$ or $\overline{r}$.

Case: $r^2 - 2 = r$. Then, $r^2 - r - 2 = 0$. The solutions are real numbers 2 and -1, which are already considered. Therefore, if the roots are complex, this equation would have real roots, which contradicts the assumption of complex roots. Therefore, there are no complex roots in this case.

Alternatively, if $r^2 - 2 = \overline{r}$. Then, since $r$ and $\overline{r}$ are roots, this would imply that $r^2 - 2 = \overline{r}$. Taking conjugate of both sides, we get $\overline{r}^2 - 2 = r$. Therefore, we have the system:

$r^2 - 2 = \overline{r}$

$\overline{r}^2 - 2 = r$

Let me denote $r = p + qi$, where p and q are real numbers. Then, $\overline{r} = p - qi$. Let me substitute into the equations.

First equation: $(p + qi)^2 - 2 = p - qi$

Compute $(p + qi)^2 = p^2 - q^2 + 2p qi$. Therefore:

$p^2 - q^2 + 2p qi - 2 = p - qi$

Equate real and imaginary parts:

Real: $p^2 - q^2 - 2 = p$

Imaginary: $2p q = -q$

Similarly, second equation: $(p - qi)^2 - 2 = p + qi$

Compute $(p - qi)^2 = p^2 - q^2 - 2p qi$. Therefore:

$p^2 - q^2 - 2p qi - 2 = p + qi$

Equate real and imaginary parts:

Real: $p^2 - q^2 - 2 = p$

Imaginary: $-2p q = q$

So, from the imaginary parts:

From first equation: $2p q = -q$. If q ≠ 0, then 2p = -1 ⇒ p = -1/2.

From second equation: -2p q = q. If q ≠ 0, then -2p = 1 ⇒ p = -1/2. So, consistent.

Therefore, p = -1/2. Then, from the real part equation:

$p^2 - q^2 - 2 = p$

Substitute p = -1/2:

$(-1/2)^2 - q^2 - 2 = -1/2$

Compute:

1/4 - q^2 - 2 = -1/2 ⇒ -q^2 - 7/4 = -1/2 ⇒ -q^2 = -1/2 + 7/4 = (-2/4 + 7/4) = 5/4 ⇒ q^2 = -5/4.

But q^2 cannot be negative. Therefore, no solution in this case. Therefore, there are no complex roots satisfying the condition. Therefore, all quadratics must have real roots. Therefore, we don't need to consider complex roots.

Therefore, our previous count of six quadratics is correct.

But let me check once more if there are any other quadratics. For example, suppose we have a quadratic with roots 2 and something else. Wait, we considered that. Similarly for other roots. Therefore, I think six is the correct answer.

But wait, hold on. Let me check if the quadratic $x^2 + x - 1$ is the only one with roots from the quadratic equation $r^2 + r - 1 = 0$. Yes, because that quadratic has two distinct roots, and we already considered that. Therefore, that's one quadratic.

Therefore, total of six quadratics. Therefore, the answer is 6. However, I need to check if the problem counts quadratics with different a and b as different even if they have the same roots. But since the problem says "quadratic equations of the form $x^2 + ax + b = 0$", each different a and b gives a different equation, even if they have the same roots. Wait, no, if two quadratics have the same roots, they are the same equation up to multiplication by a constant. But since the leading coefficient is 1, they are the same equation. Therefore, each set of roots corresponds to exactly one quadratic equation. Therefore, since we have six different sets of roots, there are six different equations.

Therefore, the answer is 6. However, I need to check again if I have considered all possibilities. Let me think again.

Wait, another way to approach this problem is to consider that the set of roots must be closed under the transformation $c \rightarrow c^2 - 2$. Since the quadratic has two roots, the set {r1, r2} must be closed under this transformation. Therefore, the possible cases are:

1. {r} where r is a fixed point (i.e., r^2 - 2 = r). These are the roots 2 and -1. So, quadratics with repeated roots 2 or -1.

2. {r, s} where r^2 - 2 = s and s^2 - 2 = r. This is the 2-cycle case, leading to the quadratic $x^2 + x - 1$.

3. {r, s} where one of them is a fixed point and the other is such that applying the transformation gives the fixed point. For example, {2, -2} because -2^2 - 2 = 2, and 2^2 - 2 = 2. Similarly, {-1, 1} because 1^2 - 2 = -1, and (-1)^2 - 2 = -1. Therefore, these are quadratics with roots {2, -2} and {-1, 1}.

Therefore, these are the four cases:

- Two fixed points (repeated roots): 2 and -1.

- One 2-cycle: {r, s} from $x^2 + x - 1$.

- Two cases where one root is a fixed point and the other maps to it: {2, -2} and {-1, 1}.

But wait, the case where one root is a fixed point and the other maps to it gives two quadratics: {2, -2} and {-1, 1}. However, the case where both roots are fixed points gives two quadratics (repeated roots). And the case of a 2-cycle gives one quadratic. Therefore, total of 2 + 1 + 2 = 5? Wait, but earlier I had six. Where is the discrepancy?

Wait, in the previous analysis, I considered:

Case 1: two quadratics (repeated roots 2 and -1)

Case 2a: one quadratic (roots 2 and -1)

Case 2b: one quadratic (2-cycle)

Case 3a: one quadratic (roots 2 and -2)

Case 4a: one quadratic (roots -1 and 1)

So, total of 2 + 1 + 1 + 1 + 1 = 6. But according to this new approach, it's 2 (repeated roots) + 1 (2-cycle) + 2 (fixed point and mapped root) = 5. Where is the difference?

Ah, because in the previous analysis, the quadratic with roots 2 and -1 is considered in both Case 1 (repeated roots) and Case 2a (distinct roots). But actually, the quadratic with roots 2 and -1 is different from the repeated roots quadratics. Therefore, in the new approach, the case where both roots are fixed points (i.e., {2, -1}) is a distinct case from the repeated roots. Therefore, the total cases are:

- Repeated roots: 2 quadratics.

- Distinct roots with both fixed points: 1 quadratic.

- 2-cycle: 1 quadratic.

- Distinct roots with one fixed point and one mapped root: 2 quadratics.

Total: 2 + 1 + 1 + 2 = 6. Yes, that matches. Therefore, the answer is 6.

Therefore, I think the correct answer is 6. However, to be thorough, let me check once more.

List of quadratics:

1. $x^2 - 4x + 4$ (repeated root 2)

2. $x^2 + 2x + 1$ (repeated root -1)

3. $x^2 - x - 2$ (roots 2 and -1)

4. $x^2 + x - 1$ (2-cycle)

5. $x^2 - 4$ (roots 2 and -2)

6. $x^2 - 1$ (roots 1 and -1)

Yes, six quadratics. Therefore, the answer is 6.

But wait, let me check if the quadratic $x^2 - x - 2$ is different from the others. Yes, because its roots are 2 and -1, which are different from the other quadratics. Therefore, all six are unique.

Therefore, I think the answer is 6. However, I recall that sometimes problems like this might have fewer solutions due to some constraints, but in this case, I think all possibilities are covered.

Therefore, my final answer is 6, so the number of such quadratic equations is \boxed{6}.

**Final Answer**
\boxed{6}

To determine the number of quadratic equations of the form $ x^2 + ax + b = 0 $ such that **whenever $ c $ is a root, $ c^2 - 2 $ is also a root**, we analyze the implications of this condition on the roots of the equation.

---

### Step 1: Understanding the Condition

Let the roots of the quadratic be $ r_1 $ and $ r_2 $. The condition implies that for each root $ c $, the value $ c^2 - 2 $ must also be a root. Since a quadratic has only two roots, this condition must be satisfied for both $ r_1 $ and $ r_2 $.

This leads to the following possibilities:

- **Case 1:** The quadratic has a **repeated root** $ r $, i.e., $ r_1 = r_2 = r $. Then, the condition becomes $ r^2 - 2 = r $, which simplifies to:
  $$
  r^2 - r - 2 = 0
  $$
  Solving this quadratic gives:
  $$
  r = 2 \quad \text{or} \quad r = -1
  $$
  So, the possible repeated root quadratics are:
  $$
  (x - 2)^2 = x^2 - 4x + 4 \quad \text{and} \quad (x + 1)^2 = x^2 + 2x + 1
  $$

- **Case 2:** The quadratic has **distinct roots** $ r_1 $ and $ r_2 $. Then, the condition implies:
  $$
  r_1^2 - 2 = r_2 \quad \text{and} \quad r_2^2 - 2 = r_1
  $$
  This system of equations leads to a quartic equation:
  $$
  r^4 - 4r^2 - r + 2 = 0
  $$
  Factoring this gives:
  $$
  (r + 1)(r - 2)(r^2 + r - 1) = 0
  $$
  The roots are:
  $$
  r = -1, \quad r = 2, \quad r = \frac{-1 \pm \sqrt{5}}{2}
  $$
  The roots $ \frac{-1 \pm \sqrt{5}}{2} $ form a **2-cycle** under the transformation $ c \mapsto c^2 - 2 $, i.e., $ r_1^2 - 2 = r_2 $ and $ r_2^2 - 2 = r_1 $. The corresponding quadratic is:
  $$
  x^2 + x - 1
  $$

- **Case 3:** One root is a fixed point (i.e., $ c^2 - 2 = c $), and the other root maps to the fixed point under the transformation. This leads to the following possibilities:

  - If one root is $ 2 $, then the other root must satisfy $ r^2 - 2 = 2 $, so $ r^2 = 4 $, giving $ r = \pm 2 $. The quadratic with roots $ 2 $ and $ -2 $ is:
    $$
    x^2 - 4
    $$

  - If one root is $ -1 $, then the other root must satisfy $ r^2 - 2 = -1 $, so $ r^2 = 1 $, giving $ r = \pm 1 $. The quadratic with roots $ -1 $ and $ 1 $ is:
    $$
    x^2 - 1
    $$

---

### Step 2: Counting the Distinct Quadratics

We now list all the distinct quadratic equations that satisfy the condition:

1. $ x^2 - 4x + 4 $ (repeated root 2)
2. $ x^2 + 2x + 1 $ (repeated root -1)
3. $ x^2 - x - 2 $ (roots 2 and -1)
4. $ x^2 + x - 1 $ (roots $ \frac{-1 \pm \sqrt{5}}{2} $)
5. $ x^2 - 4 $ (roots 2 and -2)
6. $ x^2 - 1 $ (roots 1 and -1)

Each of these quadratics has **distinct coefficients** $ a $ and $ b $, and thus represents a **unique quadratic equation** of the form $ x^2 + ax + b = 0 $.

---

### Final Answer

$$
\boxed{6}
$$

Solution:
Let the roots be $r$ and $s$ (not necessarily real).  We take the cases where $r = s$ and $r \neq s.$

Case 1: $r = s.$

Since $r$ is the only root, we must have $r^2 - 2 = r.$  Then $r^2 - r - 2 = 0,$ which factors as $(r - 2)(r + 1) = 0,$ so $r = 2$ or $r = -1.$  This leads to the quadratics $x^2 - 4x + 4$ and $x^2 + 2x + 1.$

Case 2: $r \neq s.$

Each of $r^2 - 2$ and $s^2 - 2$ must be equal to $r$ or $s.$  We have three cases:

(i) $r^2 - 2 = r$ and $s^2 - 2 = s.$

(ii) $r^2 - 2 = s$ and $s^2 - 2 = r.$

(iii) $r^2 - 2 = s^2 - 2 = r$.

In case (i), as seen from Case $r,$ $s \in \{2,-1\}.$  This leads to the quadratic $(x - 2)(x + 1) = x^2 - x - 2.$

In case (ii), $r^2 - 2 = s$ and $s^2 - 2 = r.$  Subtracting these equations, we get
\[r^2 - s^2 = s - r.\]Then $(r - s)(r + s) = s - r.$  Since $r - s \neq 0,$ we can divide both sides by $r - s,$ to get $r + s = -1.$  Adding the equations $r^2 - 2 = s$ and $s^2 - 2 = r,$ we get
\[r^2 + s^2 - 4 = r + s = -1,\]so $r^2 + s^2 = 3.$  Squaring the equation $r + s = -1,$ we get $r^2 + 2rs + s^2 = 1,$ so $2rs = -2,$ or $rs = -1.$  Thus, $r$ and $s$ are the roots of $x^2 + x - 1.$

In case (iii), $r^2 - 2 = s^2 - 2 = r.$  Then $r^2 - r - 2 = 0,$ so $r = 2$ or $r = -1.$

If $r = 2,$ then $s^2 = 4,$ so $s = -2.$  (We are assuming that $r \neq s.$)  This leads to the quadratic $(x - 2)(x + 2) = x^2 - 4.$

If $r = -1$, then $s^2 = 1,$ so $s = 1.$  This leads to the quadratic $(x + 1)(x - 1) = x^2 - 1.$

Thus, there are $\boxed{6}$ quadratic equations that work, namely $x^2 - 4x + 4,$ $x^2 + 2x + 1,$ $x^2 - x - 2,$ $x^2 + x - 1,$ $x^2 - 4,$ and $x^2 - 1.$