Question

If the roots of the equation ${px}^2+qx+r=0$, where $2p,q,2r$ are in $G.P$ are of the ${\alpha }^2,4\alpha -4$ than the value of $2p+4q+7r$ is :

• A. $82$
• B. $10$
• C. $14$
• D. $18$

Answer 1

The correct option is A $82$

Given $2p,q,2r$ are in $GP⟹q^2=4pr$

$\alpha ,4\alpha -4$ are the roots of the equation $p{x}^2+qx+r=0$

As we know that

Difference between the roots of $a{x}^2+bx+c=0$ is $\frac{\sqrt{b^2-4ac}}{a}$

$⟹{\alpha }^2-(4\alpha -4)=\frac{\sqrt{q^2-4pr}}{p}=\frac{\sqrt{4pr-4pr}}{p}=0$

$⟹{\alpha }^2-4\alpha +4=0⟹(\alpha -2{)}^2=0⟹\alpha =2$

The roots are $4,4$So the equation is $x^2-8x+16=0$

So $p=1,q=-8,r=16$

$2p+4q+7r=2-32+112=82$