Question

If $a,b$ and $c$ are 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 $\frac{1}{2}$

Explanation for the correct option:

Finding the value.

Given that, $a+2b+4c=0$

Divide whole equation by $4$,

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

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

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

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

Hence , option ‘C’ is Correct.