Question

If one root of $x^2-x-k=0$ may be the square of the other, then $k=$

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

Answer 1

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

Let the roots of the equation be $\alpha ,{\alpha }^2$
$\therefore \alpha +{\alpha }^2=1,\alpha \cdot {\alpha }^2=-k$
$\Rightarrow (\alpha +{\alpha }^2{)}^3=1,{\alpha }^3=-k$
$\Rightarrow {\alpha }^3+{\alpha }^6+3{\alpha }^3(\alpha +{\alpha }^2)=1$

now ${\alpha }^3=-k$
$\Rightarrow -k+k^2+3(-k)\left(1\right)=1$
$\Rightarrow k^2-4k-1=0$
$\Rightarrow k=\frac{4\pm \sqrt{16+4}}{2}=2\pm \sqrt{5}$