Question

Let $x_1$ and $x_2$ be the real solutions to $x^2+a x+1=0$ 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 $D \geq 0 \Rightarrow a^2-4 \geq 0 \Rightarrow a \in(-\infty,-2] \cup[2, \infty)$
Hence, only value of a is -2 .
$\therefore$ Sum of squares of all possible values of a is 4.