Question

If $x^2+ax+b=0$ and $x^2+bx+a=0$ have a common root, then

• A. $a+b+1=0$
• B. $a=b$
• C. $a+b=1$
• D. $a+b=0$

Answer 1

Answer: (A), (B)

$a=b$

Let $\alpha$ be the common root.
$\therefore {\alpha }^2+a\alpha +b=0...\left(1\right)$
and ${\alpha }^2+b\alpha +a=0...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$ we get,
$\frac{{\alpha }^2}{a^2-b^2}=\frac{\alpha }{b-a}=\frac{1}{b-a}$
$\Rightarrow {\alpha }^2=\frac{a^2-b^2}{b-a},\alpha =1$
$\Rightarrow \frac{a^2-b^2}{b-a}=1$
$\Rightarrow a^2-b^2=b-a$
$\Rightarrow (a-b)(a+b+1)=0$
$\Rightarrow a-b=0$ or $a+b+1=0$