Prove that

Solution:

Given:

$15 x^{2}-28=x$

$\Rightarrow 15 x^{2}-x-28=0$

On comparing it with $a x^{2}+b x+c=0$, we get:

$a=15, b=-1$ and $c=-28$

Discriminant $D$ is given by:

$D=\left(b^{2}-4 a c\right)$

$=(-1)^{2}-4 \times 15 \times(-28)$

$=1-(-1680)$

$=1+1680$

$=1681$

$=1681>0$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-1)+\sqrt{1681}}{2 \times 15}=\frac{1+41}{30}=\frac{42}{30}=\frac{7}{5}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-1)-\sqrt{1681}}{2 \times 15}=\frac{1-41}{30}=\frac{-40}{30}=\frac{-4}{3}$

Thus, the roots of the equation are $\frac{7}{5}$ and $\frac{-4}{3}$.