Question

The equations $x^{2}+x+a=0 $ and $x^{2}+ax+1=0$ have a common real root

• A. for no value of a
• B. for exactly one value of a
• C. for exactly two values of a
• D. for exactly three values of a

Answer 1

Answer: (B)

Let $\alpha$ be the common roots
$\therefore \alpha^{2}+\alpha +a=0$ ...(i)
and $\alpha^{2}+a \alpha+1=0$ ...(ii)
$\frac{\alpha^{2}}{1-a^{2}}=\frac{\alpha}{a-1}=\frac{1}{a-1}[a \neq 1]$
Eliminating $\alpha$, we get
$(a-1)^{2}=\left(1-a^{2}\right)(a-1)$
$\Rightarrow a-1) =1-a^{2}$
$\Rightarrow a^{2}+a-2=0$
$\Rightarrow a^{2}+2 a-a-2 =0$
$\Rightarrow a(a+2)-1(a+2) =0$
$\Rightarrow (a+2)(a-1) =0$
$\Rightarrow a =-2 .\,[\because a \neq 1]$