Trick to solve quadratic mixed with surds

Question

$$2x^2-6x-5\sqrt{x^2-3x-6}=10$$

I know that this can be solved by first taking the square root part to the right and then squaring both sides to get a quartic equation. A Google search for the equation gives that much already. But what I would like to know is whether there is any other better way for finding out the roots than just going by the conventional way.

Context: This is a question from the ‘Test of Mathematics at the 10+2 Level’ book, more specifically Q238. Actually most of the problems that I have done from the book till now have something in common: almost none of them required conventional brute force approach. So that’s why I suspect that there may be some other special way to solve the above equation.

EDIT: A fast way to factorise the equation, as below, is also welcome: $$(x-5)(x+2)(4x^2-12x-25)=0$$

Answer 1

Just a hint :)

Note that if $\sqrt{x^2 - 3x - 6} = t$ then $t^2 = x^2 - 3x -6$ or $2t^2 = 2x^2 - 6x - 12$ and $2x^2 - 6x = 2t^2 + 12$.

Hope the hint helps!

Answer 2

Notice that if we set $u = x^2 - 3x - 6$, then $2u = 2x^2 - 6x - 12$, so the equation becomes $$ 2u + 2 - 5\sqrt{u} = 0, $$ which is is equivalent to the quadratic $$ 4u^2 - 17u + 4 = 0 $$ with roots $u = \frac14$ or $u = 4$.

Now, remembering our definition of $u$, we can solve each of the quadratic equations $$ x^2 - 3x - 6 = \tfrac14 \quad\text{and}\quad x^2 - 3x - 6 = 4, $$ yielding a pair of distinct real solutions apiece.

Solutions for $x$ are $$\biggl\{ -2, \quad \frac{3 - \sqrt{34}}{2}, \quad \frac{3 + \sqrt{34}}{2}, \quad 5\, \biggr\}$$