Okay, so I need to find the number of quadratic equations of the form \(x^2 + ax + b = 0\) where whenever \(c\) is a root, \(c^2 - 2\) is also a root. Hmm, that sounds a bit tricky, but let me try to break it down step by step.

First, let's recall that a quadratic equation has at most two roots. So, let's denote the roots as \(c\) and \(d\). The problem states that if \(c\) is a root, then \(c^2 - 2\) must also be a root. Similarly, if \(d\) is a root, then \(d^2 - 2\) must also be a root. But since the equation is quadratic, there can only be two roots (counting multiplicities). So, this condition must somehow relate the roots in such a way that applying the transformation \(c \mapsto c^2 - 2\) maps one root to another, possibly itself.

Let me formalize this. Let's say the roots are \(c\) and \(c^2 - 2\). Then, since the quadratic equation has exactly two roots (assuming they are distinct or repeated), we need to check if applying the transformation again on \(c^2 - 2\) gives us back \(c\) or another root. Wait, but if we apply the transformation once, we get the other root, and applying it again should bring us back to the original root or lead to another root. However, since there are only two roots, applying the transformation twice should cycle between the roots or fix them.

So, let's assume that the roots satisfy \(r_1^2 - 2 = r_2\) and \(r_2^2 - 2 = r_1\). Then, this would create a cycle of length 2. Alternatively, maybe one root maps to itself, i.e., \(r^2 - 2 = r\). Let me explore both possibilities.

Case 1: Both roots are distinct, and each maps to the other under the transformation \(c \mapsto c^2 - 2\). So, \(r_1^2 - 2 = r_2\) and \(r_2^2 - 2 = r_1\). Then, substituting the first equation into the second, we get \((r_1^2 - 2)^2 - 2 = r_1\). Let's expand that:

\((r_1^2 - 2)^2 - 2 = r_1\)

Expanding the square:

\(r_1^4 - 4r_1^2 + 4 - 2 = r_1\)

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, a quartic equation. That's a bit complicated. Maybe there's a better way to approach this. Alternatively, maybe we can subtract the two equations. Let's subtract the first equation from the second:

\( (r_2^2 - 2) - (r_1^2 - 2) = r_1 - r_2 \)

Simplify:

\( r_2^2 - r_1^2 = r_1 - r_2 \)

Factor the left side:

\( (r_2 - r_1)(r_2 + r_1) = -(r_2 - r_1) \)

Assuming \(r_2 \neq r_1\), we can divide both sides by \(r_2 - r_1\):

\( r_2 + r_1 = -1 \)

So, the sum of the roots is -1. Wait, but for a quadratic equation \(x^2 + ax + b = 0\), the sum of the roots is \(-a\). So, that would mean \( -a = r_1 + r_2 = -1 \), so \(a = 1\). Interesting.

Now, let's use this result. We have \(r_1 + r_2 = -1\), and \(r_2 = r_1^2 - 2\). Substitute this into the sum:

\(r_1 + (r_1^2 - 2) = -1\)

Simplify:

\(r_1^2 + r_1 - 2 = -1\)

Bring all terms to one side:

\(r_1^2 + r_1 - 1 = 0\)

So, solving this quadratic equation for \(r_1\):

\(r_1 = [-1 \pm \sqrt{1 + 4}]/2 = [-1 \pm \sqrt{5}]/2\)

Therefore, the roots are \(r_1 = (-1 + \sqrt{5})/2\) and \(r_2 = (-1 - \sqrt{5})/2\), or vice versa.

Let me check if this works. Let's compute \(r_1^2 - 2\):

First, compute \(r_1^2\):

\([(-1 + \sqrt{5})/2]^2 = (1 - 2\sqrt{5} + 5)/4 = (6 - 2\sqrt{5})/4 = (3 - \sqrt{5})/2\)

Then subtract 2:

\((3 - \sqrt{5})/2 - 2 = (3 - \sqrt{5} - 4)/2 = (-1 - \sqrt{5})/2 = r_2\)

Similarly, \(r_2^2 - 2 = [(-1 - \sqrt{5})/2]^2 - 2 = (1 + 2\sqrt{5} + 5)/4 - 2 = (6 + 2\sqrt{5})/4 - 2 = (3 + \sqrt{5})/2 - 2 = (3 + \sqrt{5} - 4)/2 = (-1 + \sqrt{5})/2 = r_1\)

Perfect, so this works. So, this case gives us a quadratic equation with roots \(r_1\) and \(r_2\), sum \(-1\) and product \(r_1 r_2\). Let's compute the product:

\(r_1 r_2 = [(-1 + \sqrt{5})/2][(-1 - \sqrt{5})/2] = [1 + \sqrt{5} - \sqrt{5} - 5]/4 = (-4)/4 = -1\)

Therefore, the quadratic equation is \(x^2 - (sum)x + product = x^2 - (-1)x + (-1) = x^2 + x - 1 = 0\). So, that's one quadratic equation.

Case 2: One of the roots maps to itself. That is, there exists a root \(r\) such that \(r^2 - 2 = r\). Let's solve this equation:

\(r^2 - 2 = r\)

Bring all terms to one side:

\(r^2 - r - 2 = 0\)

Factor:

\((r - 2)(r + 1) = 0\)

So, roots are \(r = 2\) and \(r = -1\).

So, if a root is 2, then applying the transformation gives \(2^2 - 2 = 4 - 2 = 2\), which is the same root. Similarly, if a root is -1, then \((-1)^2 - 2 = 1 - 2 = -1\), which is also the same root. So, in this case, if a quadratic equation has a repeated root of 2 or -1, then the condition is satisfied, because the only root is 2 or -1, and the transformation maps it to itself.

But wait, the quadratic equation must have exactly two roots (counting multiplicity). So, if both roots are 2, then the quadratic equation is \((x - 2)^2 = x^2 - 4x + 4 = 0\). Similarly, if both roots are -1, then \((x + 1)^2 = x^2 + 2x + 1 = 0\).

But does this satisfy the condition? Let's check. For the equation with both roots 2, any root (which is 2) maps to 2, so that's fine. Similarly, for the equation with both roots -1, any root (which is -1) maps to -1. So, these are valid cases.

Additionally, we could have a quadratic equation with one root 2 and another root such that when we apply the transformation to the other root, we get 2. Wait, but if we have a root 2, then the other root must satisfy \(r^2 - 2 = 2\). Let's solve that:

\(r^2 - 2 = 2\)

\(r^2 = 4\)

\(r = 2\) or \(r = -2\)

So, if one root is 2, the other root must be 2 or -2. But if the other root is 2, then we have a repeated root, which we already considered. If the other root is -2, then we need to check if applying the transformation to -2 gives us a root. Let's see: \((-2)^2 - 2 = 4 - 2 = 2\), which is already a root. So, the quadratic equation with roots 2 and -2 satisfies the condition because applying the transformation to -2 gives 2, which is a root, and applying it to 2 gives 2, which is a root. So, this is another valid case.

Similarly, if one root is -1, then the other root must satisfy \(r^2 - 2 = -1\). Solving this:

\(r^2 - 2 = -1\)

\(r^2 = 1\)

\(r = 1\) or \(r = -1\)

So, if one root is -1, the other root must be -1 or 1. If the other root is -1, it's a repeated root, which we already have. If the other root is 1, then applying the transformation to 1 gives \(1^2 - 2 = -1\), which is a root. So, the quadratic equation with roots -1 and 1 also satisfies the condition.

Therefore, in this case, we have more quadratic equations:

1. Roots 2 and 2: equation \(x^2 - 4x + 4 = 0\)
2. Roots -1 and -1: equation \(x^2 + 2x + 1 = 0\)
3. Roots 2 and -2: equation \(x^2 - (2 + (-2))x + (2)(-2) = x^2 - 0x - 4 = x^2 - 4 = 0\)
4. Roots -1 and 1: equation \(x^2 - (-1 + 1)x + (-1)(1) = x^2 - 0x -1 = x^2 - 1 = 0\)

Wait, but let me check if these quadratics indeed satisfy the original condition.

For equation \(x^2 - 4x + 4 = 0\) (roots 2, 2): applying the transformation to 2 gives 2, which is a root. So, okay.

For equation \(x^2 + 2x + 1 = 0\) (roots -1, -1): applying the transformation to -1 gives -1, which is a root. Okay.

For equation \(x^2 - 4 = 0\) (roots 2 and -2): applying the transformation to 2 gives 2, and to -2 gives 2, which is a root. So, both roots satisfy the condition. So, this works.

For equation \(x^2 - 1 = 0\) (roots 1 and -1): applying the transformation to 1 gives -1, which is a root, and applying it to -1 gives -1, which is a root. So, this works as well.

Therefore, these four quadratics are valid. Wait, but earlier we had the case where roots are (-1 + sqrt(5))/2 and (-1 - sqrt(5))/2, giving another quadratic equation. So, in total, we have:

From Case 1: 1 quadratic equation (x² + x -1 =0)

From Case 2: 4 quadratic equations (x² -4x +4, x²+2x+1, x²-4, x²-1)

Wait, but wait: when we considered Case 2, we had roots 2 and -2, which gives x² -4 =0, and roots -1 and 1, which gives x² -1=0. But also, we had the cases where both roots are 2 or both are -1, giving (x-2)² and (x+1)². So, that's 2 more quadratics. So, in total, 1 + 4 = 5 quadratics?

Wait, let me list them:

1. x² + x -1 =0 (Case 1)
2. (x-2)² = x² -4x +4 =0
3. (x+1)² =x² +2x +1=0
4. x² -4=0 (roots 2 and -2)
5. x² -1=0 (roots 1 and -1)

Wait, but wait: x² -4=0 has roots 2 and -2, and x² -1=0 has roots 1 and -1.

So, that's 5 quadratics. But let me check if there are any overlaps. For example, is x² -4=0 the same as the quadratic from Case 1? No, because in Case 1, the roots are (-1 ± sqrt(5))/2, which are different from 2 and -2. Similarly, x² -1=0 has roots 1 and -1, which are different from the roots in Case 1.

Therefore, total of 5 quadratics. However, let me verify if each of these quadratics indeed satisfies the condition. Let's check each one:

1. Quadratic x² +x -1=0: roots are (-1 ± sqrt(5))/2. Let's check if applying the transformation to each root gives the other root. For example, take the positive root: [(-1 + sqrt(5))/2]^2 -2. Let's compute that:

First, square the root: [(-1 + sqrt(5))/2]^2 = (1 - 2sqrt(5) +5)/4 = (6 - 2sqrt(5))/4 = (3 - sqrt(5))/2.

Subtract 2: (3 - sqrt(5))/2 - 2 = (3 - sqrt(5) -4)/2 = (-1 - sqrt(5))/2, which is the other root. So, that works. So, this quadratic satisfies the condition.

2. Quadratic x² -4x +4=0: roots 2 and 2. Applying the transformation to 2 gives 2² -2=4-2=2, which is a root. So, okay.

3. Quadratic x² +2x +1=0: roots -1 and -1. Applying the transformation to -1 gives (-1)^2 -2=1-2=-1, which is a root. So, okay.

4. Quadratic x² -4=0: roots 2 and -2. Applying transformation to 2 gives 2² -2=2, which is a root. Applying to -2 gives (-2)^2 -2=4-2=2, which is a root. So, both roots satisfy the condition.

5. Quadratic x² -1=0: roots 1 and -1. Applying transformation to 1 gives 1 -2=-1, which is a root. Applying to -1 gives (-1)^2 -2=1-2=-1, which is a root. So, both roots satisfy the condition.

So, yes, all five quadratics satisfy the condition. Therefore, the answer should be 5.

Wait, but wait a second. Let me check if there are any other possibilities. For example, could there be quadratics where the roots form a cycle longer than 2? For example, three roots? But wait, the quadratic equation has only two roots. So, cycles longer than 2 are impossible. Therefore, the only possibilities are either two roots forming a cycle of length 2, or one root mapping to itself (i.e., fixed point). Therefore, the cases we considered should cover all possibilities.

But let me verify if there are any other fixed points. That is, roots that satisfy \(c^2 -2 = c\). We solved that and found c=2 and c=-1. So, those are the only fixed points. Therefore, the only quadratics that satisfy the condition are the ones with roots {2,2}, {-1,-1}, {2,-2}, {1,-1}, and the one with roots {(-1 ± sqrt(5))/2, (-1 ∓ sqrt(5))/2}. Therefore, total of five quadratics.

Therefore, the number of such quadratic equations is 5.

But let me just check if there are any other quadratics that might satisfy the condition but weren't considered. For example, suppose a quadratic equation has roots r and s, where r maps to s and s maps to r, but r ≠ s. Wait, but in the first case, we considered that, and found that leads to a quartic equation which gave us the roots with sum -1. So, that's the case when r and s are the roots of x² +x -1=0. So, that's covered.

Alternatively, suppose we have quadratics where one root maps to itself, and the other maps to the fixed root. For example, suppose one root is 2 (which maps to itself), and the other root is such that when transformed, it becomes 2. So, the other root would have to be sqrt(2 +2)=sqrt(4)=2 or -2. But if the other root is 2, then it's a repeated root, which we already considered. If the other root is -2, then the quadratic is x² -4=0, which we also considered.

Similarly, if one root is -1 (which maps to itself), and the other root is such that when transformed becomes -1. The other root must satisfy r² -2 = -1, so r²=1, so r=1 or r=-1. If the other root is -1, we have a repeated root. If the other root is 1, then the quadratic is x² -1=0, which we also considered.

Therefore, all possibilities are covered, so there are indeed 5 quadratics. Thus, the answer is 5.

**Final Answer**
The number of such quadratic equations is \boxed{5}.
Therefore, the number of quadratic equations that meet the given condition is $\boxed{5}$.
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