Question

If $alpha,beta$ are roots of $A{x}^2+Bx+C=0$ and ${\alpha }^2,{\beta }^2$ are roots of $x^2+px+q=0$, the $p$ is equal to

• A. $\frac{B^2-2AC}{A^2}$
• B. $\frac{2AC-B^2}{A^2}$
• C. $\frac{B^2-4AC}{A^2}$
• D. $\frac{4AC-B^2}{A^2}$

Answer 1

Answer: (B)

$\frac{2AC-B^2}{A^2}$

$A{x}^2+Bx+C=0$

$\alpha +\beta =\frac{-B}{A}$

$\alpha \beta =\frac{C}{A}$

Now it roots are ${\alpha }^2$ & ${\beta }^2$

Then equation will be

$(x-{\alpha }^2)(x-{\beta }^2)=0$

$x^2-x{\beta }^2-x{\alpha }^2+{\alpha }^2{\beta }^2=0$

$x^2-\left)\right({\alpha }^2+{\beta }^2)x+{\alpha }^2{\beta }^2=0$

$x^2+[-({\alpha }^2+{\beta }^2\left)\right]x+{\alpha }^2{\beta }^2=0$

$\alpha +\beta =\frac{-B}{-A}$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =\frac{B^2}{A^2}$

${\alpha }^2+{\beta }^2=\frac{B^2}{A^2}-\frac{2C}{A}$

$-({\alpha }^2+{\beta }^2)=\frac{2C}{A}-\frac{B^2}{A^2}=\frac{2AC-B^2}{A^2}$

$p=\frac{2AC-B^2}{A^2}$.