Question

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

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

Answer 1

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

Sum of roots

$\frac{k+1}{k}+\frac{k+2}{k+1}=-\frac{b}{a}\dots \dots \left(1\right)$

Product of roots

$\left(\frac{k+1}{k}\right)\left(\frac{k+2}{k+1}\right)=\frac{c}{a}\dots \dots \left(2\right)$

Eliminating $k$

Using 2nd equation

$k=\frac{2a}{c-a}$

By putting this value of k in equation(1) and rearranging the equation

$-\frac{b}{a}=\frac{a+c}{2a}+\frac{2c}{a+c}$

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

$-2ab-2bc=a^2+c^2+2ac+2ac$

$b^2-2ac=a^2+b^2+c^2+2ac+2ab+2bc$

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

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

Hence, option 'C' is correct.