Mathematics - Quadratic Equation Question with Solution | TestHub

$4x^2+2x-1=0$ $\therefore \alpha +\beta =-\frac{1}{2},\alpha \beta =-\frac{1}{4}$ Also, $4{\alpha }^2+2\alpha -1=0$ as $\alpha$ is a root and we have to prove that $\beta =4{\alpha }^3-3\alpha$. Now, $4{\alpha }^3-3\alpha =4{\alpha }^2\cdot \alpha -3\alpha$ $=\alpha (1-2\alpha )-3\alpha =-2{\alpha }^2-2\alpha$ $=-\frac{1}{2}\left[4{\alpha }^2+4\alpha \right]=-\frac{1}{2}[1-2\alpha +4\alpha ]$ $=-\frac{1}{2}(1+2\alpha )=-\frac{1}{2}-\alpha =\beta$ Now, $\alpha +\beta =-\frac{1}{2}$ [from Eq. (i)] Hence, $\beta =4{\alpha }^3-3\alpha$.