Question

Solve the following equations:
$3x^2-7+3\sqrt{3x^2-16x+21}=16x$.

Answer 1

$3x^2-7+3\sqrt{3x^2-16x+21}=16x3x^2-16x-7+3\sqrt{3x^2-16x+21}=03x^2-16x-7+28-28+3\sqrt{3x^2-16x+21}=03x^2-16x+21+3\sqrt{3x^2-16x+21}=28{\left(\sqrt{3x^2-16x+21}\right)}^2+3\sqrt{3x^2-16x+21}=28$

Put $\sqrt{3x^2-16x+21}=t$

$t^2+3t-28=0t^2+7t-4t-28=0t(t+7)-4(t+7)=0(t-4)(t+7)+0t=4,-7\sqrt{3x^2-16x+21}=t3x^2-16x+21=t^{2}3x^2-16x+21=4^{2}3x^2-16x+5=\mathrm{0........}\left(i\right)$

Also, $3x^2-16x+21={(-7)}^2$

$3x^2-16x-28=\mathrm{0........}\left(ii\right)$

Solving $\left(i\right)$
$3x^2-16x+5=03x^2-15x-x+5=03x(x-5)-1(x-5)=0(3x-1)(x-5)=0x=\frac{1}{3},5$

Solving $\left(ii\right)$

$x=\frac{16\pm \sqrt{256-4\left(3\right)(-28)}}{2\left(3\right)}=\frac{16\pm \sqrt{592}}{6}x=\frac{16\pm 2\sqrt{148}}{6}=\frac{8\pm \sqrt{148}}{3}$

So the values of $x$ are $\frac{8+\sqrt{148}}{3},\frac{8-\sqrt{148}}{3},\frac{1}{3}$ and $5$