Question

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

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

Answer 1

The correct option is A $b^2-4ac$

As $\frac{\alpha }{\alpha -1}$ and $\frac{\alpha +1}{\alpha }$ are roots of $a{x}^2+bx+c=0$ it is true that

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

and that $\frac{c}{a}=\frac{\alpha }{\alpha -1}\times \frac{\alpha +1}{\alpha }=\frac{\alpha +1}{\alpha -1}$

Substituting the sum and product of roots in standard equation, we have

$x^2-\left\{\frac{2{\alpha }^2-1}{{\alpha }^2-\alpha }\right\}x+\frac{\alpha +1}{\alpha -1}=0$

$\Rightarrow ({\alpha }^2-\alpha )x^2-(2{\alpha }^2-1)x+{\alpha }^2+\alpha =0$

on comparing this with $a{x}^2+bx+c=0$, we have

$a={\alpha }^2-\alpha$

$b=2{\alpha }^2-1$

$c={\alpha }^2+\alpha$

$\therefore (a+b+c{)}^2=1$

On comparing with given options,

$b^2-4ac=4{\alpha }^4-4{\alpha }^2+1-4({\alpha }^4-{\alpha }^2)=1=(a+b+c{)}^2$

$b^2-2ac=4{\alpha }^4-4{\alpha }^2+1-2({\alpha }^4-{\alpha }^2)=2{\alpha }^4-2{\alpha }^2+1\ne (a+b+c{)}^2$

$2b^2-ac=2(4{\alpha }^4-4{\alpha }^2+1)-({\alpha }^4-{\alpha }^2)=7{\alpha }^4-7{\alpha }^2+4\ne (a+b+c{)}^2$

Hence, option A is correct.