Question

if one roots of $x^2-x-k=0$ is square of the other, then $k=$

• A. $2\pm \sqrt{5}$
• B. $2\pm \sqrt{3}$
• C. $3\pm \sqrt{2}$
• D. $5\pm \sqrt{2}$

Answer 1

The correct option is A $2\pm \sqrt{5}$

$x^2-x-k=0$

Let $\alpha$,$\beta$ be the roots of this quadratic equation

Given that

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

We know that for a quadratic equation $a{x}^2+bx+c=0$,

$\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

Applying these results to required quadratic equation, we get

$\Rightarrow \alpha +\beta =\frac{-(-1)}{1}$

$\Rightarrow \alpha +\beta =1$ .........$\left(2\right)$

$\Rightarrow \alpha \beta =\frac{-k}{1}$

$\Rightarrow \alpha \beta =-k$ .........$\left(3\right)$

Substituting the value of $\beta$ from $\left(1\right)$ in $\left(2\right)$ and $\left(3\right)$, we get

$\Rightarrow \alpha +{\alpha }^2=1$ and $\alpha \times {\alpha }^2=-k$ .........$\left(4\right)$

$\Rightarrow {\alpha }^2+\alpha -1=0$

On solving this quadratic equation, we get

$\Rightarrow \alpha =\frac{-1\pm \sqrt{5}}{2}$

From $\left(4\right)$ we can see that

$\Rightarrow {\alpha }^3=-k$

$\Rightarrow k=-{\alpha }^3$

$\Rightarrow k=-(\frac{-1\pm \sqrt{5}}{2}{)}^3$

$\Rightarrow k=-(-2\pm \sqrt{5})$

$\Rightarrow k=2\pm \sqrt{5}$