Question

Let $a,b$ and $c$ be three real numbers, such that $a+2b+4c=0$. Then the equation $a{x}^2+bx+c=0$

• A. has both the roots complex
• B. has its roots lying within $-1<x<0$
• C. has one of the roots equal to $\frac{1}{2}$
• D. has its roots lying within $2<x<6$

Answer 1

The correct option is C has one of the roots equal to $\frac{1}{2}$

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

Then comparing it with $a{x}^2+bx+c=0$, we get

$x=\frac{1}{2}$