Question

If $x^2+x+1=0$ and $x^2+ax+b=0$ have a common root, then the minimum value of $(x-a{)}^2+2b$ is

• A. $0$
• B. $3$
• C. $2$
• D. $-2$

Answer 1

The correct option is C $2$

$x^2+x+1=0$ has non-real roots.
Also, the coefficients are real.
$\Rightarrow$ Both the roots are common.
$\therefore a=b=1$
So, $(x-a{)}^2+2b=(x-1{)}^2+2$
$\therefore$ Minimum value $=2$