Question

If the equation $a{x}^2+x+b=0$ and $x^2+bx+a=0$ have a common root and the second equation has equal roots, then $2a^2+b=$

• A. 0
• B. -1
• C. 1
• D. 2

Answer 1

The correct option is A 0

$a{x}^2+x+b=0$ and $x^2+bx+a=0$

have common root and second equation has equal root, then

in second equation

$b^2-4a=0$

$b=\pm 2\sqrt{a}$

So, root of $1^{st}$ equation

$=-\frac{b}{2}=\pm \sqrt{a}$

$1^{st}$ equation

$a{x}^2+x+b$

$a{x}^2+x+b=0$ when $x=-\frac{b}{2}$

$a\frac{b^2}{4}-\frac{b}{2}+b=0$

$b(\frac{ab}{4}-\frac{1}{2}+1)=0$

If, $b\ne 0$ so, then $a=0$

So, $2a^2+b=0$

$\frac{ab}{4}+\frac{1}{4}=0$

$ab=-2$

$2a^{3/2}=-2$

$a^{3/2}=-1$ $a\ne -1$ so, $2a^2+b=0$

Answer (A)