Question

If the quadratic equations ${ax}^2+2cx+b=0$ and ${ax}^2+2bx+c=0$ $(b\ne c)$ have a common root, then $a+4b+4c$ is equal to

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

Given,
$a{x}^2+2cx+b=0$ and $a{x}^2+2bx+c=0$ $(b\ne c)$

Let $\alpha$ be the common root .

$\Rightarrow a{\alpha }^2+2c\alpha +b=0$

$a{\alpha }^2+2b\alpha +c=0$

On subtracting we get $\alpha (2c-2b)+b-c=0$

$\Rightarrow \alpha =+\frac{1}{2}$ [$b\ne c$]

Substituting $\alpha =+\left(\frac{1}{2}\right)$ in $a{\alpha }^2+2c\alpha +b=0$ we get

$a{\left(\frac{1}{2}\right)}^2+2c\frac{1}{2}+b=0$

$\Rightarrow \frac{a}{4}+c+b=0$

$\Rightarrow a+4b+4c=0$

Hence, option 'C' is correct.