Question

If the roots fo the equation $a{x}^2+bx+c=0$, are of the form $\frac{\alpha }{\alpha -1}$ and $\frac{\alpha +1}{\alpha }$, then value of $(a+b+c{)}^2$ is

• A. $2b^2-ac$
• B. $b^2-2ac$
• C. $b^2-4ac$
• D. $4b^2-ac$

Answer 1

The correct option is A $2b^2-ac$

Given equation is $a{x}^2+bx+c=0$

Sum of the roots $\frac{\alpha }{\alpha -1}+\frac{\alpha +1}{\alpha }=-\frac{b}{a}$

$\Rightarrow \frac{{\alpha }^2+{\alpha }^2-1}{\alpha (\alpha -1)}=-\frac{b}{a}$

Product of the roots $\frac{\alpha }{\alpha -1}\times \frac{\alpha +1}{\alpha }=\frac{c}{a}$

$\Rightarrow \frac{\alpha +1}{\alpha -1}=\frac{c}{a}$

$\Rightarrow (a+b+c{)}^2={[a-\frac{{\alpha }^2+1}{\alpha (\alpha -1)}a+\frac{\alpha +1}{\alpha -1}a]}^2$

$={\left(\frac{\alpha +1}{\alpha }\right)}^2{a}^2$

Thus $2b^2-ac={\left(\frac{\alpha +1}{\alpha }\right)}^2{a}^2$ (putting value of $b$ and $c$) in terms of $a$.

$\Rightarrow (a+b+c{)}^2=2b^2-ac$