# Question d8fb4

Sep 24, 2016

The Soln. Set$= \left\{4 , - \frac{7}{2} , \frac{1 \pm \sqrt{65}}{4}\right\} .$

#### Explanation:

$\left(2 x - 7\right) \left({x}^{2} - 9\right) \left(2 x + 5\right) - 91 = 0$.

$\therefore \left\{\left(2 x - 7\right) \left(x + 3\right)\right\} \left\{\left(x - 3\right) \left(2 x + 5\right)\right\} - 91 = 0$

$\therefore \left(2 {x}^{2} - 7 x + + 6 x - 21\right) \left(2 {x}^{2} - 6 x + 5 x - 15\right) - 91 = 0$

$\therefore \left(2 {x}^{2} - x - 21\right) \left(2 {x}^{2} - x - 15\right) - 91 = 0$

$\therefore \left(y - 21\right) \left(y - 15\right) - 91 = 0 \text{, where, } y = 2 {x}^{2} - x$

$\therefore {y}^{2} - 36 y + 315 - 91 = 0$

;. y^2-36y+224=0

:. ul(y^2-28y)-ul(8y+224=0

$\therefore y \left(y - 28\right) - 8 \left(y - 28\right) = 0$

$\therefore \left(y - 28\right) \left(y - 8\right) = 0$

Returning $y$, (2x^2-x-28)(2x^2-x-8).

$\therefore \left\{\underline{2 {x}^{2} - 8 x} + \underline{7 x - 28}\right\} \left(2 {x}^{2} - x - 8\right) = 0$.

$\therefore \left\{2 x \left(x - 4\right) + 7 \left(x - 4\right)\right\} \left(2 {x}^{2} - x - 8\right) = 0$

$\therefore \left(x - 4\right) \left(2 x + 7\right) \left(2 {x}^{2} - x - 8\right) = 0$

$\therefore x = 4 , \mathmr{and} , x = - \frac{7}{2} , \mathmr{and} , 2 {x}^{2} - x - 8 = 0$
,
To find the zeroes of $\left(2 {x}^{2} - x - 8\right) = 0$, we use the Quadratic Formula & get the Zeroes : $\frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(2\right) \left(- 8\right)}}{2 \cdot 2} = \frac{1 \pm \sqrt{65}}{4}$.

Thus, the Soln. Set$= \left\{4 , - \frac{7}{2} , \frac{1 \pm \sqrt{65}}{4}\right\} .$

Enjoy Maths.!