Help with radical equation

Question

Please, help me to solve this equation. No advanced math should be needed.

$$ 3x^2 - 4x + \sqrt{3x^2 - 4x - 6} = 18 $$

I'm clueless. It should be simple.

Answer 1

Hint : set $u:= \sqrt{3 x^2-4x-6}$ then your equation becomes $u^2+6+u=18$ or $$u^2+u-12=0$$

Answer 2

Put $u = 3x^2 - 4x$. Then the equation becomes $$ u + \sqrt{u - 6} = 18.$$ Once you solve for $u$ in the auxillary equation, you are left with a quadratic.

Answer 3

No advanced math is needed.

• Denote $u = \sqrt{3 x^2 - 4x -6}$.
• Write the original equation in terms of $u$. You should get a quadratic equation.
• Solve it.
• For each root $u_\ast$ found, solve $3 x^2 - 4x -6 = u_\ast^2$.
• Verify your solutions against the original equation and weed out extraneous solutions.