Solve for $x \in \mathbb{R}$

$$4x^2(x+2) +3(2x^2-4x-3)\sqrt{4x+3} +6x = 0$$

I tried taking square by isolating the radical, but the resultant equation couldn't be solved.

Any help will be appreciated.
Thanks.

• I suspect you are expected to make an educated guess here by checking values of $x$ which cause $4x+3$ to be a perfect square. But then $x$ cannot be an integer since all odd squares are of the form $4n+1$. So $\frac{3}{2}$ would be a good first guess since $4\left(\frac{3}{2}\right)+3=9$. Checking this value verifies that it is a solution. To see if there are other real solutions would require eliminating the radical and applying perhaps Descartes Rule of signs and other tests. – John Wayland Bales May 4 '16 at 4:52
• Confession: I actually tried $\frac{1}{4}$ first but it was not a solution. You would also check $-\frac{1}{2}$. – John Wayland Bales May 4 '16 at 4:57

Let $X=4x+3$. Then, $$3(2x^2-X)\sqrt{X}=-2x(2x^2+X)$$ Squaring the both sides gives $$9(2x^2-X)^2X=4x^2(2x^2+X)^2,$$ i.e. $$4x^4X-4x^2X^2+9X^3-16x^6+16x^4X-36x^2X^2=0$$ $$X(4x^4-4x^2X+9X^2)-4x^2(4x^4-4x^2X+9X^2)=0$$ $$(X-4x^2)(4x^4-4x^2X+9X^2)=0$$ since $9X^2=X^2+8X^2$ $$(X-4x^2)(4x^4-4x^2X+X^2+8X^2)=0$$ to have $$(X-4x^2)((2x^2-X)^2+8X^2)=0$$ finally, $$(-4x^2+4x+3)((2 x^2-4 x-3)^2+8 (4 x+3)^2)=0$$ and so $$x=\frac 32,-\frac 12.$$ The former is sufficient while the latter isn't. So, $\color{red}{x=\frac 32}$ is the only solution.