Question

Solve the following equations:
$(2x-7)(x^2-9)(2x+5)=91$.

Answer 1

$(2x-7)(x^2-9)(2x+5)=91(2x-7)(x+3)(x-3)(2x+5)=91\left\{\right(2x-7\left)\right(x+3\left)\right\}\left\{\right(x-3\left)\right(2x+5\left)\right\}=91(2x^2+6x-7x-21)(2x^2+5x-6x-15)=91(2x^2-x-21)(2x^2-x-15)=91$

Put $2x^2-x=t$

$(t-21)(t-15)=91t^2-15t-21t+315-91=0t^2-36t+224=0t^2-28t-8t+224=0t(t-28)-8(t-28)=0(t-8)(t-28)=0t=8,282x^2-x=82x^2-x-8=\mathrm{0.....}\left(i\right)2x^2-x=282x^2-x-28=\mathrm{0.......}\left(ii\right)$

Solving $\left(i\right)$

$2x^2-x-8=0$

Using quadratic formula

$x=\frac{1\pm \sqrt{1-4\left(2\right)(-8)}}{2\left(2\right)}=\frac{1\pm \sqrt{65}}{4}$

Solving $\left(ii\right)$

$2x^2-x-28=02x^2-8x+7x-28=02x(x-4)+7(x-4)=0(2x+7)(x-4)=0x=-\frac{7}{2},4$

So the values of $x$ are $\frac{1+\sqrt{65}}{4},\frac{1-\sqrt{65}}{4},-\frac{7}{2}$ and $4$