Question

If one root of the equation $a{x}^2+bx+c=0,a,b,cϵR,$ is the cube of other, then $\frac{b^2-2ac}{a+c}$ is equal to

• A. $\pm \sqrt{ac}$
• B. $\pm \sqrt{(}bc)$
• C. $\pm \frac{c^2}{a}$
• D. $\pm \frac{c^2}{b}$

Answer 1

The correct option is A $\pm \sqrt{ac}$

Let the roots be $\alpha$and${\alpha }^3$
$\therefore \alpha +{\alpha }^3=-\frac{b}{a}$ and ${\alpha }^4=\frac{c}{a}$
$\Rightarrow {\alpha }^2(1+{\alpha }^2{)}^2=\frac{b^2}{a^2}$
$\Rightarrow {\alpha }^2+2{\alpha }^4+{\alpha }^6=\frac{b^2}{a^2}$
$\Rightarrow \sqrt{\frac{c}{a}}+\frac{2c}{a}+\frac{c}{a}\sqrt{\frac{c}{a}}=\frac{b^2}{a^2}$
$\Rightarrow \sqrt{\frac{c}{a}}(1+\frac{c}{a})=\frac{b^2-2ac}{a^2}$
$\Rightarrow \frac{c}{a}[\frac{a+c}{a}{]}^2=\frac{(b^2-2ac{)}^2}{a^4}$
$\Rightarrow ac=(\frac{b^2-2ac}{a+c}{)}^2$
$\therefore \frac{b^2-2ac}{a+c}=\pm \sqrt{ac}$