Question

If $\alpha$ and $\beta$ are roots of $a{x}^2+bx+c=0$ then the equation whose roots are ${\alpha }^2$ and ${\beta }^2$ is

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

Answer 1

The correct option is A $a^2{x}^2-(b^2-2ac)x+c^2=0$


If $\alpha$ and $\beta$ are roots of
$a{x}^2+bx+c=0$ then,
$\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

The equation whose roots are ${\alpha }^2$ and ${\beta }^2$ will be,
$x^2-({\alpha }^2+{\beta }^2)x+{\alpha }^2{\beta }^2=0\cdots \left(1\right)$
Now,
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$\Rightarrow {\alpha }^2+{\beta }^2={(-\frac{b}{a})}^2-2\left(\frac{c}{a}\right)$
$\Rightarrow {\alpha }^2+{\beta }^2=\frac{b^2-2ac}{a^2}$
${\alpha }^2{\beta }^2={\left(\frac{c}{a}\right)}^2=\frac{c^2}{a^2}$

Substituting the value of
${\alpha }^2+{\beta }^2$ and $\alpha \beta$ in equation (1)

$\Rightarrow x^2-\left(\frac{b^2-2ac}{a^2}\right)x+\frac{c^2}{a^2}=0$

Hence the required quadratic equation is,
$a^2{x}^2-(b^2-2ac)x+c^2=0$