Question

The number of pairs (a, b) of real numbers, such that whenever $\alpha$ is a root of the equation $x^2 + ax + b = 0$, $\alpha^2 - 2$ is also a root of this equation, is:

• A. 6
• B. 8
• C. 4
• D. 2

Answer 1

Answer: (A)

6

Answer 2

Let the given quadratic equation be $x^2 + ax + b = 0$. Let its roots be $r_1$ and $r_2$. According to Vieta's formulas: $r_1 + r_2 = -a$ $r_1 r_2 = b$

The problem states that if $\alpha$ is a root of the equation, then $\alpha^2 - 2$ is also a root. This means the set of roots $\{r_1, r_2\}$ must be closed under the transformation $f(x) = x^2 - 2$.

We consider different cases for the roots:

Case 1: The roots are equal, i.e., $r_1 = r_2 = r$. Since $r$ is a root, $f(r) = r^2 - 2$ must also be a root. As there's only one distinct root $r$, it must be that $r^2 - 2 = r$. $r^2 - r - 2 = 0$ Factoring the quadratic equation: $(r-2)(r+1) = 0$. This yields two possible values for $r$: $r=2$ or $r=-1$.

• If $r=2$: The roots are $(2, 2)$. $-a = 2+2 = 4 \implies a = -4$ $b = 2 \times 2 = 4$ This gives the pair $(a, b) = (-4, 4)$. Check: $x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0$. The only root is $2$. If $\alpha=2$, $\alpha^2-2 = 2^2-2 = 2$, which is a root. This pair is valid.

• If $r=-1$: The roots are $(-1, -1)$. $-a = (-1)+(-1) = -2 \implies a = 2$ $b = (-1) \times (-1) = 1$ This gives the pair $(a, b) = (2, 1)$. Check: $x^2 + 2x + 1 = 0 \implies (x+1)^2 = 0$. The only root is $-1$. If $\alpha=-1$, $\alpha^2-2 = (-1)^2-2 = -1$, which is a root. This pair is valid.

Case 2: The roots are distinct, i.e., $r_1 \neq r_2$. The condition $f(S) \subseteq S$ implies that either: a) Both roots are fixed points of $f(x)$: $r_1 = f(r_1)$ and $r_2 = f(r_2)$. b) The roots form a 2-cycle under $f(x)$: $r_1 = f(r_2)$ and $r_2 = f(r_1)$. c) One root is a fixed point, and the other maps to that fixed point.

Let's analyze these subcases:

Subcase 2a: Both roots are fixed points ($r_1 = f(r_1)$ and $r_2 = f(r_2)$). This means both $r_1$ and $r_2$ are solutions to $x^2 - x - 2 = 0$. The solutions are $x=2$ and $x=-1$. Since the roots are distinct, $\{r_1, r_2\} = \{2, -1\}$. $-a = 2+(-1) = 1 \implies a = -1$ $b = 2 \times (-1) = -2$ This gives the pair $(a, b) = (-1, -2)$. Check: $x^2 - x - 2 = 0$. Roots are $2$ and $-1$. If $\alpha=2$, $\alpha^2-2 = 2^2-2 = 2$, which is a root. If $\alpha=-1$, $\alpha^2-2 = (-1)^2-2 = -1$, which is a root. This pair is valid.

Subcase 2b: The roots form a 2-cycle ($r_1 = f(r_2)$ and $r_2 = f(r_1)$). We have the system of equations:

1. $r_1 = r_2^2 - 2$
2. $r_2 = r_1^2 - 2$ Subtracting (2) from (1): $r_1 - r_2 = r_2^2 - r_1^2 = (r_2 - r_1)(r_2 + r_1)$. Since $r_1 \neq r_2$, we can divide by $(r_1 - r_2)$: $1 = -(r_1 + r_2)$, so $r_1 + r_2 = -1$. Substitute $r_2 = -1 - r_1$ into $r_1 = r_2^2 - 2$: $r_1 = (-1 - r_1)^2 - 2$ $r_1 = (1 + 2r_1 + r_1^2) - 2$ $r_1^2 + r_1 - 1 = 0$ The roots are $r_{1,2} = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$. These are distinct real roots. $-a = r_1 + r_2 = -1 \implies a = 1$ $b = r_1 r_2 = \left(\frac{-1 + \sqrt{5}}{2}\right)\left(\frac{-1 - \sqrt{5}}{2}\right) = \frac{(-1)^2 - (\sqrt{5})^2}{4} = \frac{1 - 5}{4} = -1$ This gives the pair $(a, b) = (1, -1)$. Check: $x^2 + x - 1 = 0$. The roots $r_{1,2}$ satisfy $x^2 = 1-x$. $f(r_1) = r_1^2 - 2 = (1-r_1) - 2 = -1-r_1$. Since $r_1+r_2=-1$, $-1-r_1 = r_2$. So $f(r_1)=r_2$. Similarly, $f(r_2)=r_1$. This pair is valid.

Subcase 2c: One root is a fixed point, and the other maps to that fixed point. Let $r_1$ be a fixed point, so $r_1 \in \{2, -1\}$. Let $r_2$ not be a fixed point ($r_2 \neq f(r_2)$). The condition $f(r_2) \in \{r_1, r_2\}$ implies $f(r_2) = r_1$ (since $r_2 \neq f(r_2)$).

• If $r_1 = 2$: Then $f(r_2) = 2 \implies r_2^2 - 2 = 2 \implies r_2^2 = 4 \implies r_2 = \pm 2$. Since $r_1 \neq r_2$, $r_2$ must be $-2$. So, the roots are $\{2, -2\}$. $-a = 2+(-2) = 0 \implies a = 0$ $b = 2 \times (-2) = -4$ This gives the pair $(a, b) = (0, -4)$. Check: $x^2 - 4 = 0$. Roots are $2$ and $-2$. If $\alpha=2$, $\alpha^2-2 = 2^2-2 = 2$, which is a root. If $\alpha=-2$, $\alpha^2-2 = (-2)^2-2 = 2$, which is a root. This pair is valid.

• If $r_1 = -1$: Then $f(r_2) = -1 \implies r_2^2 - 2 = -1 \implies r_2^2 = 1 \implies r_2 = \pm 1$. Since $r_1 \neq r_2$, $r_2$ must be $1$. So, the roots are $\{-1, 1\}$. $-a = -1+1 = 0 \implies a = 0$ $b = (-1) \times 1 = -1$ This gives the pair $(a, b) = (0, -1)$. Check: $x^2 - 1 = 0$. Roots are $1$ and $-1$. If $\alpha=1$, $\alpha^2-2 = 1^2-2 = -1$, which is a root. If $\alpha=-1$, $\alpha^2-2 = (-1)^2-2 = -1$, which is a root. This pair is valid.

Combining all valid pairs $(a, b)$:

1. $(-4, 4)$
2. $(2, 1)$
3. $(-1, -2)$
4. $(1, -1)$
5. $(0, -4)$
6. $(0, -1)$

There are 6 distinct pairs of real numbers $(a, b)$ that satisfy the given condition.

The final answer is $\boxed{6}$.

Explanation of the solution: The problem requires finding pairs $(a,b)$ such that if $\alpha$ is a root of $x^2+ax+b=0$, then $\alpha^2-2$ is also a root. Let the roots be $r_1, r_2$. The condition implies the set of roots $\{r_1, r_2\}$ is closed under the transformation $f(x)=x^2-2$.

1. Equal roots ($r_1=r_2=r$): $r$ must be a fixed point of $f(x)$, so $r^2-2=r \implies r^2-r-2=0 \implies (r-2)(r+1)=0$.
   • If $r=2$: $a=-(2+2)=-4, b=2 \times 2=4$. Pair: $(-4,4)$.
   • If $r=-1$: $a=-(-1-1)=2, b=(-1)(-1)=1$. Pair: $(2,1)$.
2. Distinct roots ($r_1 \neq r_2$):
   • Both roots are fixed points: $r_1, r_2$ are distinct solutions to $x^2-x-2=0$. So $\{r_1, r_2\} = \{2, -1\}$. $a=-(2-1)=-1, b=2 \times (-1)=-2$. Pair: $(-1,-2)$.
   • Roots form a 2-cycle: $r_1=f(r_2)$ and $r_2=f(r_1)$. This leads to $r_1^2-2=r_2$ and $r_2^2-2=r_1$. Subtracting gives $r_1-r_2 = -(r_1-r_2)(r_1+r_2)$, so $1=-(r_1+r_2) \implies r_1+r_2=-1$. Substituting $r_2=-1-r_1$ into $r_1=r_2^2-2$ yields $r_1^2+r_1-1=0$. $a=-(r_1+r_2)=-(-1)=1, b=r_1r_2=-1$. Pair: $(1,-1)$.
   • One root fixed, other maps to it: Let $r_1$ be a fixed point ($r_1 \in \{2,-1\}$) and $f(r_2)=r_1$ (since $r_2$ is not a fixed point).
     • If $r_1=2$: $r_2^2-2=2 \implies r_2^2=4 \implies r_2=\pm 2$. Since $r_1 \neq r_2$, $r_2=-2$. Roots $\{2,-2\}$. $a=-(2-2)=0, b=2 \times (-2)=-4$. Pair: $(0,-4)$.
     • If $r_1=-1$: $r_2^2-2=-1 \implies r_2^2=1 \implies r_2=\pm 1$. Since $r_1 \neq r_2$, $r_2=1$. Roots $\{-1,1\}$. $a=-(-1+1)=0, b=(-1) \times 1=-1$. Pair: $(0,-1)$.

All 6 pairs are distinct.

Answer: 6