Question

If one of the roots of the quadratic equation $x^2-x-1=0$ is $\alpha$, then its other root is

• A. ${\alpha }^2-3\alpha$
• B. ${\alpha }^3-2\alpha$
• C. ${\alpha }^2-2\alpha$
• D. ${\alpha }^3-3\alpha$

Answer 1

The correct option is D ${\alpha }^3-3\alpha$

Let $\beta$ be other root of equation $x^2-x-1=0$

$\therefore \alpha +\beta =-\frac{-1}{1}=1$

$\Rightarrow \beta =1-\alpha \cdots \left(i\right)$

$\because \alpha$ is a root of, $x^2-x-1=0$

$\therefore {\alpha }^2-\alpha -1=0$

$\Rightarrow {\alpha }^2=\alpha +1\cdots \left(ii\right)$

Now to find the correct expression, let's do hit and trial method.
${\alpha }^2-3\alpha =\alpha +1-3\alpha =1-2\alpha =\beta -\alpha \ne \beta$

${\alpha }^3-2\alpha ={\alpha }^2\cdot \alpha +1-2\alpha =\alpha (\alpha +1)-2\alpha ={\alpha }^2+\alpha -2\alpha =\alpha +1-\alpha =1\ne \beta$

${\alpha }^2-2\alpha =\alpha +1-2\alpha =1-\alpha =-\beta \ne \beta$

${\alpha }^3-3\alpha =\alpha .{\alpha }^2-3\alpha$
$=\alpha (\alpha +1)-3\alpha$
$={\alpha }^2+\alpha -3\alpha$
$=\alpha +1+\alpha -3\alpha$
$=1-\alpha =\beta$

Hence other root $\beta ={\alpha }^3-3\alpha$