Question

Let $x_1$ and $x_2$ be the real solutions to $x^2+ax+1=0$ for some real number a. If $x_1+1$ and $x_2+1$ are solutions of the equation $x^2-a^{2}x+a^2=0$, then the sum of squares of all possible values of $a$ is

• A. 1
• B. 4
• C. 5
• D. 7

Answer 1

Answer: (B)

$x_1+x_2=-a$
$x_1{x}_2=1$
Also $x_1+x_2+2=a^2$
(1) & (3) $\Rightarrow a^2+a-2=0\Rightarrow (a+2)(a-1)=0\Rightarrow a=1$ or $a=-2$
Also, $\left(x_1+1\right)\left(x_2+1\right)=a^2$
$x_1{x}_2+\left(x_1+x_2\right)+1=a^2\Rightarrow 1-a+1=a^2\Rightarrow a^2+a-2=0\Rightarrow (a+2)(a-1)=0$
$\Rightarrow a=1\text{ or }a=-2$
Also, for real roots $D\ge 0\Rightarrow a^2-4\ge 0\Rightarrow a\in (-\infty ,-2]\cup [2,\infty )$
Hence, only value of a is -2 .
$\therefore$ Sum of squares of all possible values of a is 4.