Question

Solve the following equations: $x^2+2\sqrt{x^2+6x}=24-6x$.

Answer 1

$x^2+2\sqrt{x^2+6x}=24-6x{x}^2+6x+2\sqrt{x^2+6x}=24{\left(\sqrt{x^2+6x}\right)}^2+2\sqrt{x^2+6x}=24$

Put $\sqrt{x^2+6x}=t$

$t^2+2t=24t^2+2t-24=0t^2+6t-4t-24=0t(t+6)-4(t+6)=0(t-4)(t+6)=0t=4,-6$

$\sqrt{x^2+6x}=t{x}^2+6x=t^2{x}^2+6x=4^2{x}^2+6x-16=\mathrm{0......}\left(i\right)x^2+6x={(-6)}^2{x}^2+6x-36=\mathrm{0.......}\left(ii\right)$

Solving $\left(i\right)$
$x^2+6x-16=0x^2+8x-2x-16=0x(x+8)-2(x+8)=0(x-2)(x+8)=0x=2,-8$

Solving $\left(ii\right)$
$x^2+6x-36=0$

Using quadratic formula

$x=\frac{-6\pm \sqrt{36-4\left(1\right)(-36)}}{2}=\frac{-6\pm \sqrt{180}}{2}x=\frac{-6\pm 6\sqrt{5}}{2}=-3\pm 3\sqrt{5}$

So the values of $x$ are $-3+3\sqrt{5},-3-3\sqrt{5},-8$ and $2$