Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f\left(x\right)=x^2-(\sqrt{5}-1)x-(\sqrt{5}+1)$, then the value of $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ is ____________.

• A. $3+\sqrt{5}$
• B. $3-\sqrt{5}$
• C. $\sqrt{5}-3$
• D. $-3-\sqrt{5}$

Answer 1

The correct option is B $3-\sqrt{5}$

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ $=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}$

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

$={\left(\frac{\alpha +\beta }{\alpha \beta }\right)}^2-\frac{2}{\alpha \beta }$ .....(1)

SInce $\alpha$, $\beta$ are the zeros of the quadratic polynomial, there are also the roots of the quadratic equation $x^2-(\sqrt{5}-1)x-(\sqrt{5}+1)=0$

Sum of the roots $=\alpha +\beta =-\frac{b}{a}=\sqrt{5}-1$ .....(2)

Product of the roots $=\alpha \beta =\frac{c}{a}=-(\sqrt{5}+1)$ .....(3)

Substituting (2),(3) in (1), we get

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ $=$ ${\left(\frac{\sqrt{5}-1}{-(\sqrt{5}+1)}\right)}^2-\frac{2}{-(\sqrt{5}+1)}$

$=\frac{(\sqrt{5}-1{)}^2+2(\sqrt{5}+1)}{(\sqrt{5}+1{)}^2}$ .....(Taking LCM )

$=\frac{(5-2\sqrt{5}+1)+2(\sqrt{5}+1)}{(5+2\sqrt{5}+1{)}^2}$

$=\frac{8}{(6+2\sqrt{5})}$

$=\frac{8}{2(3+\sqrt{5})}$

$=\frac{4}{(3+\sqrt{5})}$

$=\frac{4(3-\sqrt{5})}{(3^2-(\sqrt{5}{)}^2)}$ .....(Multiplying & Dividing by $3-\sqrt{5}$)

$=\frac{4(3-\sqrt{5})}{(9-5)}$

$=3-\sqrt{5}$