Question

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

• A. $\frac{1}{2}$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is D $0$

Let $\alpha$ be the common root.
$\Rightarrow a{\alpha }^2+2c\alpha +b=0$ and $a{\alpha }^2+2b\alpha +c=0$
$\Rightarrow 2\alpha (c-b)+(b-c)=0$
$\Rightarrow \alpha =\frac{1}{2}(\because b\ne c)$

So, $a{\left(\frac{1}{2}\right)}^2+2c\left(\frac{1}{2}\right)+b=0$
$\Rightarrow a+4b+4c=0$