Question

If $\alpha ,\beta$ are the roots of $2x^2+3x-1=0,$ then the equation whose roots are $\frac{1}{{\alpha }^2},\frac{1}{{\beta }^2}$ is

• A. $x^2-5x+4=0$
• B. $x^2-13x+4=0$
• C. $x^2-3x-2=0$
• D. $x^2+5x+4=0$

Answer 1

The correct option is B $x^2-13x+4=0$

Given equation $2x^2+3x-1=0$ has roots as $\alpha ,\beta$
Let $y=\frac{1}{x^2}$
$\Rightarrow x=\frac{1}{\sqrt{y}}$
Replace $x$ by $\frac{1}{\sqrt{y}}$ in the given equation, we get
$\frac{2}{y}+\frac{3}{\sqrt{y}}-1=0\Rightarrow (\frac{2}{y}-1)=-\frac{3}{\sqrt{y}}$
Squaring on both sides,
${(\frac{2}{y}-1)}^2=\frac{9}{y}\Rightarrow \frac{4}{y^2}+1-\frac{4}{y}=\frac{9}{y}\Rightarrow y^2-13y+4=0$
So, the equation whose roots are $\frac{1}{{\alpha }^2},\frac{1}{{\beta }^2}$ is $x^2-13x+4=0$