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. $\sum a^2$
• C. $b^2-2ac$
• D. $b^2-4ac$

Answer 1

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

Sum of the roots$=$$\frac{k+1}{k}+\frac{k+2}{k+1}=\frac{-b}{a}\to \left(1\right)$

Product of roots$=$$\frac{k+1}{k}\times \frac{k+2}{k+1}=\frac{c}{a}\to \left(2\right)⟹\frac{k+2}{k}=\frac{c}{a}or\frac{2}{k}=\frac{c}{a}-1ork=\frac{2a}{c-a}$

putting $k=\frac{2a}{c-a}$ in eqn(1).

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

adding $b^2$ to both sides we get,

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