<Question>
Find the number of quadratic equations of the form $x^2 + ax + b = 0,$ such that whenever $c$ is a root of the equation, $c^2 - 2$ is also a root of the equation.
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this problem. So, the question is asking for the number of quadratic equations of the form \(x^2 + ax + b = 0\) such that whenever \(c\) is a root, then \(c^2 - 2\) is also a root. Hmm, interesting. So, first, let me parse this condition carefully. The quadratic equation has roots \(c\) and maybe another root, right? But the condition says that if \(c\) is a root, then \(c^2 - 2\) must also be a root. So, this implies that the roots must come in pairs where each root's image under the transformation \(c \mapsto c^2 - 2\) is also a root.
Since it's a quadratic equation, there can be at most two distinct roots. So, either both roots are fixed points of this transformation, meaning that applying \(c^2 - 2\) to each root gives the same root back, or the roots form a cycle of length 2 under this transformation.

Let me write down the quadratic equation as \(x^2 + ax + b = 0\). Let the roots be \(r_1\) and \(r_2\). Then, by Vieta's formula, we have:

\(r_1 + r_2 = -a\)

\(r_1 r_2 = b\)

So, our goal is to find all possible pairs \((a, b)\) such that whenever \(c\) is a root, \(c^2 - 2\) is also a root.

<prompt_1.1> Restate the problem and recall Vieta's formulas for a quadratic equation. </prompt_1.1>

First, let's consider the case where the quadratic has a repeated root. That is, \(r_1 = r_2 = r\). Then, the equation becomes \((x - r)^2 = x^2 - 2rx + r^2 = 0\). Comparing with \(x^2 + ax + b = 0\), we have \(a = -2r\) and \(b = r^2\).

<prompt_1.2> Consider the case of a repeated root and determine the conditions for the root to satisfy the given property. </prompt_1.2>
-----

Now, moving on to the case where the quadratic has two distinct roots. So, \(r_1 \neq r_2\). Then, the condition is that if \(r_1\) is a root, then \(r_1^2 - 2\) must also be a root. But since there are only two roots, \(r_1^2 - 2\) must be either \(r_1\) or \(r_2\). Similarly, applying the same transformation to \(r_2\), we must have \(r_2^2 - 2\) is either \(r_2\) or \(r_1\).

<prompt_1.3> Consider the case of two distinct roots and determine the conditions for the roots to satisfy the given property. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let's parse this condition carefully. The quadratic equation has roots \(c\) and maybe another root, right? But the condition says that if \(c\) is a root, then \(c^2 - 2\) must also be a root. So, this implies that the roots must come in pairs where each root's image under the transformation \(c \mapsto c^2 - 2\) is also a root.
Since it's a quadratic equation, there can be at most two distinct roots. So, either both roots are fixed points of this transformation, meaning that applying \(c^2 - 2\) to each root gives the same root back, or the roots form a cycle of length 2 under this transformation.

Let me write down the quadratic equation as \(x^2 + ax + b = 0\). Let the roots be \(r_1\) and \(r_2\). Then, by Vieta's formula, we have:

\(r_1 + r_2 = -a\)

\(r_1 r_2 = b\)

So, our goal is to find all possible pairs \((a, b)\) such that whenever \(c\) is a root, \(c^2 - 2\) is also a root.
</execution_1.1>

<execution_1.2>
Now, the condition that if \(c\) is a root, then \(c^2 - 2\) is also a root. Since both roots are the same, we need \(r^2 - 2 = r\). So, solving \(r^2 - r - 2 = 0\). Let's factor this quadratic: \(r^2 - r - 2 = (r - 2)(r + 1) = 0\), so roots are \(r = 2\) and \(r = -1\). Therefore, for these values of \(r\), the repeated root equations would satisfy the given condition.

So, if \(r = 2\), then \(a = -2*2 = -4\), and \(b = 2^2 = 4\). So, the equation is \(x^2 -4x +4 =0\).

If \(r = -1\), then \(a = -2*(-1) = 2\), and \(b = (-1)^2 = 1\). So, the equation is \(x^2 +2x +1 =0\).
</execution_1.2>

<execution_1.3>
Therefore, there are two possibilities:

1. Each root is a fixed point of the transformation, i.e., \(r_i^2 - 2 = r_i\) for \(i = 1, 2\).
2. The roots form a 2-cycle, meaning \(r_1^2 - 2 = r_2\) and \(r_2^2 - 2 = r_1\).

We need to explore both cases.

Case 1: Fixed points. For each root \(r_i\), we have \(r_i^2 - r_i - 2 = 0\). Solving this quadratic equation:

\(r = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3] / 2\), so \(r = 2\) or \(r = -1\).

Therefore, the possible roots are 2 and -1. But since we are in the case of two distinct roots, both roots must be fixed points. So, the quadratic must have roots 2 and -1. Let's check if this satisfies the condition.

If the roots are 2 and -1, then applying the transformation to each root:

For 2: \(2^2 - 2 = 4 - 2 = 2\), which is a root.

For -1: \((-1)^2 - 2 = 1 - 2 = -1\), which is also a root.

So, this quadratic does satisfy the condition. Therefore, this is a valid case.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
But wait, are these the only possibilities? Let me check.

Suppose the equation has two distinct roots \(r_1\) and \(r_2\). Then, according to the problem's condition, if \(r_1\) is a root, then \(r_1^2 - 2\) must also be a root. But since there are only two roots, \(r_1^2 - 2\) must be either \(r_1\) or \(r_2\).

<prompt_2.1> Verify the condition for a repeated root by checking if the transformation maps the root to itself. </prompt_2.1>
-----

Based on execution_1.3:
Case 2: 2-cycle. Here, \(r_1^2 - 2 = r_2\) and \(r_2^2 - 2 = r_1\). Let's set up these equations:

From the first equation: \(r_1^2 - 2 = r_2\)

<prompt_2.2> Solve for possible 2-cycles where the roots form a cycle under the transformation. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Similarly, applying the transformation to \(r_2\), \(r_2^2 - 2\) must be either \(r_2\) or \(r_1\). So, there are a few possibilities here.

Subcase 1: \(r_1^2 - 2 = r_1\) and \(r_2^2 - 2 = r_2\). Then both roots are fixed points.

Subcase 2: \(r_1^2 - 2 = r_2\) and \(r_2^2 - 2 = r_1\). Then, they form a 2-cycle.

Subcase 3: \(r_1^2 - 2 = r_1\) and \(r_2^2 - 2 = r_1\). So, one root is a fixed point, and the other maps to the fixed point.

Similarly, Subcase 4: \(r_1^2 - 2 = r_2\) and \(r_2^2 - 2 = r_2\). Here, one root maps to the other, which is a fixed point.

But maybe Subcases 3 and 4 are included in the above cases. Let's analyze each subcase.

Subcase 1: Both roots satisfy \(r^2 - 2 = r\), which we already solved. The roots are 2 and -1. So, this gives us the quadratic with roots 2 and -1, as above.
</execution_2.1>

<execution_2.2>
From the second equation: \(r_2^2 - 2 = r_1\)

Substitute \(r_2 = r_1^2 - 2\) into the second equation:

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

Expand the left side:

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

Simplify:

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

Bring all terms to one side:

\(r_1^4 - 4r_1^2 - r_1 + 2 = 0\)

So, we need to solve this quartic equation. Let's see if we can factor it.

Looking for rational roots using Rational Root Theorem. Possible rational roots are ±1, ±2.

Test r=1: 1 -4 -1 +2 = -2 ≠0

Test r= -1: 1 -4 +1 +2=0. Wait, 1 -4 is -3, -3 +1 is -2, -2 +2=0. So, r= -1 is a root.

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

Using polynomial division or synthetic division:

Divide \(r^4 -4r^2 -r +2\) by (r +1):

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

Using synthetic division with root -1:

Bring down 1.

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

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

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

Multiply by -1: 2*(-1)= -2. Add to last coefficient: 2 + (-2)=0. So, remainder is 0.

Therefore, the quartic factors as (r +1)(r^3 - r^2 -3r +2)=0.

Now, let's factor the cubic \(r^3 - r^2 -3r +2\). Again, try rational roots: ±1, ±2.

Test r=1: 1 -1 -3 +2= -1 ≠0

Test r=2: 8 -4 -6 +2=0. So, r=2 is a root.

Factor out (r -2) from the cubic:

Using synthetic division:

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

Divide by (r -2), so root 2:

Bring down 1.

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

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

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

Thus, the cubic factors as (r -2)(r^2 + r -1)=0.

Therefore, the quartic factors as (r +1)(r -2)(r^2 + r -1)=0.

So, the roots are r = -1, 2, and roots of \(r^2 + r -1 =0\), which are \(r = [-1 ± sqrt(1 +4)]/2 = [-1 ± sqrt(5)]/2\).

Therefore, the quartic equation \(r_1^4 -4r_1^2 -r_1 +2=0\) has roots at r= -1, 2, \((-1 + sqrt(5))/2\), and \((-1 - sqrt(5))/2\).

But in Case 2, we are considering a 2-cycle where \(r_1\) and \(r_2\) are distinct roots. So, if \(r_1\) is one of the roots of the quartic, then \(r_2 = r_1^2 - 2\), and we need to check if this forms a 2-cycle.

Let's check each root:

1. If \(r_1 = -1\), then \(r_2 = (-1)^2 -2 = 1 -2 = -1\). So, \(r_2 = -1\), which is the same as \(r_1\). So, this would not form a 2-cycle; it's a fixed point. So, discard this.

2. If \(r_1 = 2\), then \(r_2 = 2^2 -2 = 4 -2 = 2\). Again, same as \(r_1\). So, another fixed point. So, discard.

3. If \(r_1 = [ -1 + sqrt(5) ] / 2\), let's compute \(r_2 = r_1^2 -2\):

First, compute \(r_1^2\):

\(r_1 = \frac{ -1 + \sqrt{5} }{ 2 }\)

So, \(r_1^2 = \left( \frac{ -1 + \sqrt{5} }{ 2 } \right)^2 = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}\)

Then, \(r_2 = r_1^2 -2 = \frac{3 - \sqrt{5}}{2} - 2 = \frac{3 - \sqrt{5} -4}{2} = \frac{ -1 - \sqrt{5} }{ 2 }\)

Which is the other root. So, \(r_2 = \frac{ -1 - \sqrt{5} }{ 2 }\). Then, check if applying the transformation to \(r_2\) gives back \(r_1\):

Compute \(r_2^2 -2\):

\(r_2 = \frac{ -1 - \sqrt{5} }{ 2 }\)

\(r_2^2 = \left( \frac{ -1 - \sqrt{5} }{ 2 } \right)^2 = \frac{1 + 2\sqrt{5} +5}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}\)

Then, \(r_2^2 -2 = \frac{3 + \sqrt{5}}{2} -2 = \frac{3 + \sqrt{5} -4}{2} = \frac{ -1 + \sqrt{5} }{ 2 } = r_1\)

So, indeed, \(r_1\) and \(r_2\) form a 2-cycle. Therefore, this is a valid case. So, the quadratic equation with roots \(r_1\) and \(r_2\) would satisfy the given condition.

Similarly, if we start with \(r_1 = \frac{ -1 - \sqrt{5} }{ 2 }\), we get \(r_2 = \frac{ -1 + \sqrt{5} }{ 2 }\), which is the other root. So, this 2-cycle is valid.

Therefore, in Case 2, we have one valid 2-cycle, leading to a quadratic equation with roots \(r_1\) and \(r_2\) as above.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Then, \(a = -(2 + (-1)) = -1\), and \(b = 2*(-1) = -2\). So, the equation is \(x^2 - x -2 = 0\), which factors as \((x - 2)(x + 1) = 0\), indeed having roots 2 and -1.

But wait, is this the only case where the roots are fixed points? Let's see. The fixed points are 2 and -1, so if the quadratic has these as roots, then they satisfy the condition.

But are there any other quadratics with two distinct roots where each root is mapped to itself?

<prompt_3.1> Determine if there are any other quadratics with two distinct roots where each root is mapped to itself. </prompt_3.1>

Now, moving on to Subcase 2: One root is a fixed point, and the other maps to the fixed point.

<prompt_3.2> Analyze Subcase 2 where one root is a fixed point and the other maps to the fixed point. </prompt_3.2>

Subcase 3: One root is a fixed point, and the other maps to itself. Wait, but if one root is a fixed point and the other is also a fixed point, that's the same as Subcase 1. So maybe Subcases 3 and 4 are already covered.

Alternatively, perhaps Subcases 3 and 4 are already included in Subcase 1. Let me check.

<prompt_3.3> Re-evaluate Subcases 3 and 4 to ensure they are included in Subcase 1. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
From the quadratic equation, the roots would have to be 2 and -1, as those are the only fixed points. So, yes, that seems unique. So, only one such quadratic with distinct roots.

But wait, hold on. Suppose there are other roots where \(c^2 - 2 = c\), but we already found all roots of that equation. The equation \(r^2 - r - 2 = 0\) has only two roots: 2 and -1. So, those are the only possible fixed points.
</execution_3.1>

<execution_3.2>
Wait, but if \(r_1\) is a fixed point (so \(r_1 = 2\) or \(r_1 = -1\)), and \(r_2\) maps to \(r_1\), then \(r_2^2 - 2 = r_1\). So, let's suppose \(r_1 = 2\). Then, \(r_2^2 - 2 = 2\) implies \(r_2^2 = 4\), so \(r_2 = 2\) or \(r_2 = -2\). But since the roots are \(2\) and \(-2\), let's check if this satisfies the condition.

If the roots are 2 and -2, then applying the transformation to 2 gives \(2^2 - 2 = 2\), which is a root. Applying the transformation to -2 gives \((-2)^2 - 2 = 4 - 2 = 2\), which is also a root. So, the roots would be 2 and 2? Wait, no. Wait, if the roots are 2 and -2, but the transformation of -2 is 2, which is a root, and transformation of 2 is 2, which is a root. So, actually, the roots would be 2 and -2. But does this quadratic equation satisfy the condition? Let's check.

The quadratic equation with roots 2 and -2 is \(x^2 - (2 + (-2))x + (2)(-2) = x^2 + 0x -4 = x^2 -4\). So, \(a = 0\), \(b = -4\). Let's check the condition. If c is a root, then c^2 -2 must be a root.

For c=2: 2^2 -2 = 4 -2 = 2, which is a root.

For c=-2: (-2)^2 -2 = 4 -2 = 2, which is a root.

So, yes, both 2 and -2 are roots, and their transformations are also roots. Therefore, this quadratic equation does satisfy the condition. Therefore, this is another valid case.

Similarly, if \(r_1 = -1\), then \(r_2^2 -2 = -1\), so \(r_2^2 = 1\), so \(r_2 = 1\) or \(r_2 = -1\). So, the roots would be -1 and 1. Let's check this quadratic.

The quadratic equation with roots -1 and 1 is \(x^2 - ( -1 +1 )x + (-1)(1) = x^2 +0x -1 = x^2 -1\). So, \(a=0\), \(b=-1\). Check the condition:

For c=1: 1^2 -2 = -1, which is a root.

For c=-1: (-1)^2 -2 = -1, which is a root.

So, yes, this works. Therefore, another valid quadratic equation is \(x^2 -1 =0\), with \(a=0\), \(b=-1\).

Therefore, in Subcase 2, we have two possibilities:

Either the roots are 2 and -2 (equation \(x^2 -4 =0\)), or the roots are -1 and 1 (equation \(x^2 -1 =0\)). So, these are two more quadratics.
</execution_3.2>

<execution_3.3>
If \(r_1\) is a fixed point (so \(r_1^2 - 2 = r_1\)), and \(r_2^2 - 2 = r_1\), then \(r_2\) maps to \(r_1\), which is a fixed point. But if \(r_1\) is a fixed point, then \(r_2\) must be such that \(r_2^2 - 2 = r_1\). However, since \(r_1\) is a fixed point, \(r_1^2 - 2 = r_1\), so \(r_2^2 - 2 = r_1\) implies \(r_2^2 = r_1 + 2\). But since \(r_1\) is a root, and the quadratic is \(x^2 + ax + b = 0\), we have \(r_1 + r_2 = -a\) and \(r_1 r_2 = b\). Maybe this can lead to another possibility. Wait, but if \(r_1\) is a fixed point, then \(r_1\) is either 2 or -1. Let's see.

Suppose \(r_1 = 2\), which is a fixed point. Then, if \(r_2^2 - 2 = 2\), so \(r_2^2 = 4\), so \(r_2 = 2\) or \(r_2 = -2\). But since the quadratic has roots 2 and -2, let's check if this satisfies the condition. Then, the quadratic would be \((x - 2)(x + 2) = x^2 -4\), so \(a = 0\), \(b = -4\). Then, the roots are 2 and -2. Applying the transformation to 2: 2^2 - 2 = 2, which is a root. Applying to -2: (-2)^2 - 2 = 4 - 2 = 2, which is also a root. So, this quadratic also satisfies the condition. Wait, so this is another case where one root is a fixed point (2), and the other root (-2) maps to the fixed point 2. So, this is a different quadratic equation. Therefore, this is a valid case. Similarly, if the fixed point is -1, let's check.

Suppose \(r_1 = -1\), a fixed point. Then, \(r_2^2 - 2 = -1\), so \(r_2^2 = 1\), so \(r_2 = 1\) or \(r_2 = -1\). If the roots are -1 and 1, then the quadratic is \((x +1)(x -1) = x^2 -1\), so \(a = 0\), \(b = -1\). Check the condition: roots are -1 and 1. Applying transformation to -1: (-1)^2 -2 = -1, which is a root. Applying to 1: 1^2 -2 = -1, which is a root. So, this quadratic also satisfies the condition. Therefore, these are additional quadratics.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_3.2, execution_3.3:
So, summarizing Case 2 (two distinct roots):

Subcase 1: Both roots are fixed points (2 and -1), leading to equation \(x^2 - x -2 =0\).

Subcase 2: One root is a fixed point, the other maps to it (2 and -2, leading to \(x^2 -4 =0\); or -1 and 1, leading to \(x^2 -1 =0\)).

Therefore, in Case 2, we have three quadratics:

1. \(x^2 - x -2 =0\) (roots 2 and -1)

<prompt_4.1> List and describe the quadratics found in Case 2. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
2. \(x^2 -4 =0\) (roots 2 and -2)

3. \(x^2 -1 =0\) (roots -1 and 1)

Wait, but hold on. Wait, the quadratic \(x^2 -4 =0\) has roots 2 and -2. But according to the condition, if c is a root, then c^2 -2 must be a root. For c=2: 2^2 -2 = 2, which is a root. For c=-2: (-2)^2 -2 = 2, which is a root. So, yes, this works. Similarly for the others.

But wait, the quadratic \(x^2 -1 =0\) has roots 1 and -1. Applying the transformation:

For c=1: 1^2 -2 = -1, which is a root.

For c=-1: (-1)^2 -2 = -1, which is a root. So, yes, this works.

Therefore, in total, in Case 2, we have three quadratics:

- \(x^2 -x -2 =0\) (distinct roots 2 and -1)

- \(x^2 -4 =0\) (distinct roots 2 and -2)

- \(x^2 -1 =0\) (distinct roots -1 and 1)

Wait, but earlier, when we considered the 2-cycle, we found roots \((-1 ± sqrt(5))/2\), but those are distinct from these. So, these are separate quadratics.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Therefore, combining all cases:

From Case 1 (repeated roots):

- \(x^2 -4x +4 =0\) (double root 2)

- \(x^2 +2x +1 =0\) (double root -1)

From Case 2 (distinct roots):

- \(x^2 -x -2 =0\) (roots 2 and -1)

- \(x^2 -4 =0\) (roots 2 and -2)

- \(x^2 -1 =0\) (roots -1 and 1)

So, total of 6 quadratics. But wait, let's check if these quadratics are distinct.

<prompt_5.1> Verify the distinctness of the quadratics found. </prompt_5.1>

But wait, let me check if there are any other possibilities. For example, suppose there is a quadratic with roots forming a 2-cycle with other numbers. But we already considered the 2-cycle case, which gave roots involving sqrt(5). However, those roots lead to a quadratic equation with those roots. Let's see.

<prompt_5.2> Explore other possibilities for roots and verify if they lead to valid quadratics. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Let's list them:

1. \(x^2 -4x +4 =0\)

2. \(x^2 +2x +1 =0\)

3. \(x^2 -x -2 =0\)

4. \(x^2 -4 =0\)

5. \(x^2 -1 =0\)

Wait, that's five quadratics. Wait, no:

Wait, in Case 1, two quadratics. In Case 2, three quadratics. So, total 2 +3=5. But let me check each:

1. \(x^2 -4x +4 =0\) (double root 2)

2. \(x^2 +2x +1 =0\) (double root -1)

3. \(x^2 -x -2 =0\) (roots 2 and -1)

4. \(x^2 -4 =0\) (roots 2 and -2)

5. \(x^2 -1 =0\) (roots -1 and 1)

Yes, five distinct quadratics.
</execution_5.1>

<execution_5.2>
Earlier, we found that if we have a 2-cycle, the roots are \((-1 ± sqrt(5))/2\). Let's compute those roots:

Let \(r = [ -1 + sqrt(5) ] / 2 ≈ ( -1 + 2.236 ) / 2 ≈ 1.236 / 2 ≈ 0.618\), and the other root is \([ -1 - sqrt(5) ] / 2 ≈ ( -1 - 2.236 ) / 2 ≈ -3.236 / 2 ≈ -1.618\).

So, these roots are approximately 0.618 and -1.618. Let's check if applying the transformation to each gives the other root.

For r ≈ 0.618: \(r^2 - 2 ≈ (0.618)^2 - 2 ≈ 0.618*0.618 ≈ 0.381 - 2 ≈ -1.619\), which is approximately the other root. Then, applying the transformation to -1.618: \((-1.618)^2 - 2 ≈ 2.618 - 2 ≈ 0.618\). So, yes, this forms a 2-cycle. Therefore, this quadratic equation would have these two roots, and thus would satisfy the given condition. Therefore, this is another quadratic equation.

Similarly, the sum of roots is \(r_1 + r_2 = [ (-1 + sqrt(5))/2 + (-1 - sqrt(5))/2 ] = (-2)/2 = -1\), so \(a = -(-1) = 1\). The product is \(r_1 r_2 = [ (-1 + sqrt(5))/2 ] [ (-1 - sqrt(5))/2 ] = [ ( (-1)^2 - (sqrt(5))^2 ) ] / 4 = (1 -5)/4 = (-4)/4 = -1\), so \(b = -1\). Therefore, the quadratic equation is \(x^2 + x -1 =0\). Wait, let's confirm:

Sum of roots: \(r_1 + r_2 = [ (-1 + sqrt(5))/2 + (-1 - sqrt(5))/2 ] = (-2)/2 = -1\), so \(-a = -1\) implies \(a =1\). Product of roots: \([ (-1)^2 - (sqrt(5))^2 ] /4 = (1 -5)/4 = -1\), so \(b = -1\). Therefore, the quadratic is \(x^2 + x -1 =0\). Let's verify the condition:

For root \(r_1 ≈0.618\): \(r_1^2 -2 ≈ -1.618\), which is \(r_2\). For root \(r_2 ≈-1.618\): \(r_2^2 -2 ≈0.618\), which is \(r_1\). So, yes, this quadratic satisfies the condition. Therefore, this is another quadratic equation.

Therefore, this is a valid case. Therefore, this quadratic equation must be included.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
Therefore, adding this to our previous count, total quadratics are 6.

But wait, let's check if there's another quadratic from the 2-cycle. Wait, the 2-cycle gives one quadratic. So, adding that to the previous five, total six.

<prompt_6.1> Verify the distinctness of all quadratics found. </prompt_6.1>

But wait, let's check the discriminant for each quadratic to ensure they are distinct:

<prompt_6.2> Verify the discriminants of the quadratics to ensure they are distinct. </prompt_6.2>

But wait, let's check if there are any other quadratics we might have missed. For example, suppose there's a quadratic with both roots fixed points and part of a 2-cycle. But since fixed points are unique (only 2 and -1), and the 2-cycle involves other roots, which we already considered.

<prompt_6.3> Perform a final verification to ensure no other quadratics are missed. </prompt_6.3>
</Plan_6>

<Execution_6>
<execution_6.1>
Wait, no. Wait, in Case 2, we had three quadratics, but one of them (\(x^2 -x -2 =0\)) actually has roots 2 and -1, which are also fixed points. So, maybe there's overlap between Case 1 and Case 2.

Wait, yes. The quadratic \(x^2 -x -2 =0\) has roots 2 and -1, which are fixed points. So, this quadratic is included in both Case 1 and Case 2. Therefore, we need to check for overlaps.

Similarly, the quadratic \(x^2 -1 =0\) has roots -1 and 1. The root -1 is a fixed point, but 1 is not a fixed point. However, 1 maps to -1, which is a root. So, this is a distinct quadratic from the one in Case 1.

Similarly, \(x^2 -4 =0\) has roots 2 and -2. The root 2 is a fixed point, but -2 is not. However, -2 maps to 2, which is a root. So, this is another distinct quadratic.

Therefore, the total quadratics are:

From Case 1:

1. \(x^2 -4x +4 =0\) (double root 2)

2. \(x^2 +2x +1 =0\) (double root -1)

From Case 2:

3. \(x^2 -x -2 =0\) (roots 2 and -1)

4. \(x^2 -4 =0\) (roots 2 and -2)

5. \(x^2 -1 =0\) (roots -1 and 1)

6. \(x^2 +x -1 =0\) (roots \((-1 ± sqrt(5))/2\))

Wait, but wait, the quadratic \(x^2 +x -1 =0\) is separate. So, total six quadratics.
</execution_6.1>

<execution_6.2>
1. \(x^2 -4x +4 =0\): Discriminant \(16 -16 =0\), repeated root.

2. \(x^2 +2x +1 =0\): Discriminant \(4 -4 =0\), repeated root.

3. \(x^2 -x -2 =0\): Discriminant \(1 +8 =9\), distinct roots.

4. \(x^2 -4 =0\): Discriminant \(0\), distinct roots.

5. \(x^2 -1 =0\): Discriminant \(0\), distinct roots.

6. \(x^2 +x -1 =0\): Discriminant \(1 +4 =5\), distinct roots.

So, all these quadratics have distinct roots except the first two, which have a repeated root. Therefore, all six quadratics are distinct.
</execution_6.2>

<execution_6.3>
Therefore, the total number of quadratics is 6.

Wait, but let's check if the quadratic \(x^2 +x -1 =0\) is distinct from the others. Its roots are different from all previous roots, so yes, it's a distinct quadratic. Therefore, total of six quadratics.
</execution_6.3>
</Execution_6>

<Plan_7>
Based on execution_6.3:
But let me verify once again:

<prompt_7.1> Verify each quadratic and its roots to confirm they satisfy the condition. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
1. \(x^2 -4x +4 =0\): roots 2 (double). Check.

2. \(x^2 +2x +1 =0\): roots -1 (double). Check.

3. \(x^2 -x -2 =0\): roots 2 and -1. Check.

4. \(x^2 -4 =0\): roots 2 and -2. Check.

5. \(x^2 -1 =0\): roots -1 and 1. Check.

6. \(x^2 +x -1 =0\): roots \((\frac{-1 ± sqrt(5)}{2})\). Check.

Yes, all satisfy the condition. Therefore, total of six quadratics.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
But wait, let me check if there are any other quadratics. For example, suppose there's a quadratic with roots that are part of a 2-cycle but also include fixed points. But since fixed points are already considered in Case 1, and the 2-cycle is separate, I don't think so.

<prompt_8.1> Perform a final verification to ensure no other quadratics are missed. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Alternatively, perhaps there's a quadratic where one root is a fixed point and the other is part of a 2-cycle. For example, if one root is 2 (a fixed point) and the other root is part of a 2-cycle with another root. But in that case, the other root would have to map to the first root. Let's see.

Suppose we have a quadratic with roots 2 and \(r\), where \(r^2 -2 =2\), so \(r^2=4\), so \(r=2\) or \(r=-2\). So, this brings us back to the quadratic \(x^2 -4 =0\), which we already considered. So, that's already included.

Similarly, if we have a root that is part of a 2-cycle with a fixed point. For example, suppose we have a root \(r\) such that \(r^2 -2 =2\), which gives \(r=2\) or \(r=-2\). So, that's already covered.

Alternatively, if a root \(r\) satisfies \(r^2 -2 = s\), and \(s^2 -2 = r\), forming a 2-cycle. Which is exactly the case we considered earlier, leading to the roots involving \(\frac{-1 ± sqrt(5)}{2}\). So, that's already accounted for.

Therefore, all possible quadratics are accounted for. Therefore, the total number is 6. But let me check once more if there are any other quadratics.

Wait, another thought: suppose the quadratic has roots that are both part of a 2-cycle, but not the one we found. For example, suppose there's another pair of roots that form a 2-cycle, but we didn't consider. However, when we solved for the quartic equation, we found only one 2-cycle in addition to the fixed points. So, the only possible 2-cycles are the one we found. Therefore, only one additional quadratic from the 2-cycle.

Therefore, total quadratics are:

From fixed points:

1. \(x^2 -4x +4 =0\)

2. \(x^2 +2x +1 =0\)

From 2-cycles:

3. \(x^2 -x -2 =0\)

4. \(x^2 -4 =0\)

5. \(x^2 -1 =0\)

6. \(x^2 +x -1 =0\)

Yes, that's six quadratics. Therefore, the answer should be 6.

But wait, let me check if the quadratic \(x^2 +x -1 =0\) is indeed distinct and satisfies the condition. Yes, as we saw earlier, its roots are \(\frac{-1 ± sqrt(5)}{2}\), and applying the transformation cycles them. So, that's valid.

Therefore, the total number of such quadratics is 6.

**Final Answer**
The number of such quadratic equations is \boxed{6}.
</execution_8.1>
</Execution_8>

<Final_answer>
Therefore, the final answer is $\boxed{6}$.
</Final_answer>
</think>