Question

Solve the following equations:
$x(2x+1)(x-2)(2x-3)=63$.

Answer 1

$x(2x+1)(x-2)(2x-3)=63\left\{x\right(2x-3\left)\right\}\left\{\right(2x+1\left)\right(x-2\left)\right\}=63({2x}^2-3x)(2x^2-4x++x-2)=63(2x^2-3x)(2x^2-3x-2)=63$

Put $2x^2-3x=t$

$t(t-2)=63t^2-2t-63=0t^2-9t+7t-63=0t(t-9)+7(t-9)=0(t+7)(t-9)=0t=-7,92x^2-3x=t2x^2-3x=-72x^2-3x+7=\mathrm{0.......}\left(i\right)2x^2-3x=92x^2-3x-9=\mathrm{0........}\left(ii\right)$

Solving $\left(i\right)$
$2x^2-3x+7=0x=\frac{3\pm \sqrt{9-4\left(2\right)\left(7\right)}}{2\left(2\right)}=\frac{3\pm \sqrt{9-56}}{4}x=\frac{3\pm \sqrt{-47}}{4}$

Solving $\left(ii\right)$

$2x^2-3x-9=02x^2-6x+3x-9=02x(x-3)+3(x-3)=0(2x+3)(x-3)=0x=-\frac{3}{2},3$

So the values of $x$ are $\frac{3+\sqrt{-47}}{4},\frac{3-\sqrt{-47}}{4},-\frac{3}{2}$ and $3$