Question

Solve the following equations:
$8+9\sqrt{(3x-1)(x-2)}=3x^2-7x$.

Answer 1

$8+9\sqrt{(3x-1)(x-2)}=3x^2-7x8+9\sqrt{3x^2-6x-x+2}=3x^2-7x8+9\sqrt{3x^2-7x+2}=3x^2-7x$

Adding $2$ on both the sides

$10+9\sqrt{3x^2-7x+2}=3x^2-7x+210+9\sqrt{3x^2-7x+2}={\left(\sqrt{3x^2-7x+2}\right)}^2\sqrt{3x^2-7x+2}=t10+9t=t^2{t}^2-9t-10=0t^2-10t+t-10=0t(t-10)+1(t-10)=0$

$(t+1)(t-10)=0$

$t=-1,10$

Then, $\sqrt{3x^2-7x+2}=t$

$3x^2-7x+2=t^2$

$3x^2-7x+2={(-1)}^2$

$3x^2-7x+1=\mathrm{0.........}\left(i\right)$

Using quadratic formula

$x=\frac{7\pm \sqrt{49-4\left(3\right)\left(1\right)}}{2\left(3\right)}=\frac{7\pm \sqrt{37}}{6}$

Also, $3x^2-7x+2={10}^2$

$3x^2-7x-98=03x^2-21x+14x-98=03x(x-7)+14(x-7)=0(3x+14)(x-7)=0x=-\frac{14}{3},7$

So the values of $x$ are $\frac{7-\sqrt{37}}{6},\frac{7+\sqrt{37}}{6},-\frac{14}{3}$ and $7$