Question

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

• A. $x^2+(2b-a^2)x+a^2=0$
• B. $x^2+(2b-a^2)x+b^2=0$
• C. $b{x}^2+(2b-a^2)x+b=0$
• D. $x^2+(a^2-2b)x+b^2=0$

Answer 1

The correct option is A

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

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

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

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