Question

If $\alpha ,\beta$ are roots of the equation $a{x}^2+bx+c=0,$ then the quadratic equation whose roots are $\frac{1}{(a\alpha +b{)}^2},\frac{1}{(a\beta +b{)}^2}$, is

• A. $\left(a^2\right)x^2+(2ac-b^2)x+1=0$
• B. $\left(c^2\right)x^2+(2ac-b^2)x+1=0$
• C. $\left(a^2{c}^2\right)x^2+(2ac-b^2)x+1=0$
• D. $\left(a^2{c}^2\right)x^2+(2ac+b^2)x+1=0$

Answer 1

The correct option is C $\left(a^2{c}^2\right)x^2+(2ac-b^2)x+1=0$

We know that $\alpha ,\beta$ are the roots of
$a{x}^2+bx+c=0\cdots \left(1\right)$
Now, $a{\alpha }^2+b\alpha +c=0$
$\Rightarrow \alpha (a\alpha +b)+c=0$
$\Rightarrow (a\alpha +b)=-\frac{c}{\alpha },(a\beta +b)=-\frac{c}{\beta }$
Therefore, $\frac{1}{(a\alpha +b{)}^2}=\frac{{\alpha }^2}{c^2}$
and $\frac{1}{(a\beta +b{)}^2}=\frac{{\beta }^2}{c^2}$

Let $y=\frac{x^2}{c^2}$, where $x=\alpha ,\beta$
$c^{2}y=x^2\Rightarrow x=c\sqrt{y}$
Putting this in equation $\left(1\right)$,
$a(c\sqrt{y}{)}^2+b(c\sqrt{y})+c=0\Rightarrow a{c}^{2}y+bc\sqrt{y}+c=0\Rightarrow (a{c}^{2}y+c{)}^2=(-bc\sqrt{y}{)}^2\Rightarrow a^2{c}^4{y}^2+(2a{c}^3-b^2{c}^2)y+c^2=0\Rightarrow (a^2{c}^2)y^2+(2ac-b^2)y+1=0$

Hence, the required quadratic equation is,
$\left(a^2{c}^2\right)x^2+(2ac-b^2)x+1=0$