Question

Suppose $a,bϵRanda\ne 0,b\ne 0$. Let $\alpha ,\beta$ be the roots of $x^2+ax+b=0$ . Find the equation whose roots are ${\alpha }^2$, ${\beta }^2$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We want to find the equation whose roots are ${\alpha }^2$, ${\beta }^2$. We can say if x is a root of the equation $x^2+ax+b=0$ , we want to find an equation whose root is $x^2$. Let it be y.

$\Rightarrow y=x^2$

$\Rightarrow x=\sqrt{y}$

Replace x by $\sqrt{y}$ in $x^2$ + ax + b = 0 (Because x satisfies the equation, $\sqrt{y}$ also satisfies the equation)

$\Rightarrow$ ${\sqrt{y}}^2$ + a $\sqrt{y}$ + b = 0

$\Rightarrow$ y + b = - a$\sqrt{y}$

$\Rightarrow$ ${(y+b)}^2$ = $a^2$ y

$\Rightarrow$ $y^2$ + 2by + $b^2$ = $a^2$ y

$y^2$ + (2b - $a^2$) y + $b^2$ = 0

This is the equation whose roots are y or $x^2$ or ${\alpha }^2$, ${\beta }^2$

Since the changing of variable does not affect an equation, we can write it as

$x^2$ + (2b - $a^2$) x + $b^2$ = 0

2nd method:

${\alpha }^2$, ${\beta }^2$ are roots of the required equation

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

$\alpha +\beta =-a$

$\alpha \beta =b$

$\Rightarrow$ ${\alpha }^2$ + ${\beta }^2$ = $a^2$ - 2b [Sum of roots]

${\alpha }^2$ ${\beta }^2$ = $(\alpha \beta {)}^2$ [Product of roots]

= $b^2$

$\Rightarrow$ The equation is

$x^2$ - ( $a^2$ - 2b)x + $b^2=0$

$x^2$ + (2b - $a^2$)x + $b^2=0$