Solve the equation $4x^3+13x^2-14x=3-\sqrt{15x+9}$ I ran into this problem:
$$4x^3+13x^2-14x=3-\sqrt{15x+9}$$
It makes no sense to square it. I thought it was necessary to make a replacement. What kind of substitution?
First of all, the root gets in the way, so you have to make a substitution that removes the root!
I was able to find such a substitution
$$x=\frac{9}{15}\cos 2t\Rightarrow \sqrt{15x+9}=3\sqrt{2}\cos t$$
We don't have to put the module, because the restrictions allow us to do so
After we got rid of the root, then comes the second problem. How do we solve this equation?
$$\frac{108}{125}\cos^32 t+\frac{117}{25}\cos^22t-\frac{42}{5}\cos 2t=3-3\sqrt{2}\cos t$$
 A: Rewrite the equation and square both sides:
$$3-4x^3-13x^2+14x=\sqrt{15x+9}$$
$$(3-4x^3-13x^2+14x)^2=16x^6+104x^5+57x^4-388x^3+118x^2+84x+9 $$
You have to solve:
$$16x^6+104x^5+57x^4-388x^3+118x^2+69x=0 $$
which can be factored in the following way:
$$x(4x-3)(4x^4+29x^3+36x^2-70x-23)=0
 $$
A: The LHS of the equation
$$4x^3+13x^2-14x=3-\sqrt{15x+9}$$
equals $0$ when $x=0$ for which the RHS also equals $0$.
So, here, I would add $ax$ to the both sides (each side still equals $0$ when $x=0$) since after having
$$4x^3+13x^2-14x+ax=ax+3-\sqrt{15x+9}$$
$$4x^3+13x^2-14x+ax=\frac{(ax+3)^2-(15x+9)}{ax+3+\sqrt{15x+9}}$$
$$x(4x^2+13x-14+a)=\frac{x(a^2x+6a-15)}{ax+3+\sqrt{15x+9}}\tag1$$
it may be possible to have something like
$$x(x-c)(\text{something positive})=0$$
if the both sides of $(1)$ have a common factor $(x-c)$.
This means that the solution of $a^2x+6a-15=0$ has to be a solution of $4x^2+13x-14+a=0$, which means that $a$ has to satisfy
$$4\bigg(\frac{-6a+15}{a^2}\bigg)^2+13\bigg(\frac{-6a+15}{a^2}\bigg)-14+a=0$$
i.e.
$$\frac{a^5 - 14 a^4 - 78 a^3 + 339 a^2 - 720 a + 900}{a^4} = 0$$
Fortunately the rational root theorem works to get $a=2$.

Solution :
$$\begin{align}&4x^3+13x^2-14x=3-\sqrt{15x+9}
\\\\&\iff 4x^3+13x^2-14x+2x=2x+3-\sqrt{15x+9}
\\\\&\iff x(4x-3)(x+4)=\frac{(2x+3)^2-(15x+9)}{2x+3+\sqrt{15x+9}}
\\\\&\iff x(4x-3)(x+4)=\frac{x(4x-3)}{2x+3+\sqrt{15x+9}}
\\\\&\iff x(4x-3)\bigg(x+4-\frac{1}{2x+3+\sqrt{15x+9}}\bigg)=0
\\\\&\iff x(4x-3)\bigg(x+1+3-\frac{1}{2x+3+\sqrt{15x+9}}\bigg)=0
\\\\&\iff x(4x-3)\bigg(\underbrace{x+1+\frac{6(x+1)+2+3\sqrt{15x+9}}{2(x+1)+1+\sqrt{15x+9}}}_{\text{positive since $x\geqslant -9/15\gt -1$}}\bigg)=0
\\\\&\iff x=0,\frac 34\end{align}$$
A: There is a way to construct a possible solution using only conjugates , without applying squaring.
To get something useful, we want to multiply both side of the equation by the conjugate $3+\sqrt {15x+9}\neq 0\thinspace $ :
$$\begin{align}x\left(3+\sqrt {15x+9}\right)\left(4x^2+13x-14\right)=-15x\end{align}$$
Since $3+\sqrt {15x+9}\neq 0$ for all $x\geq -\frac 35$, this implies that $x_1=0$ is a solution. Therefore, to find other possible roots, we can proceed by dividing both side of the equation by $x\thinspace (x\neq 0\thinspace)$ :
$$
\begin{align}\left(3+\sqrt {15x+9}\right)\left(4x^2+13x-14\right)=-15\end{align}
$$
By rearranging the left-hand side of the equation, we have :
$$
\begin{align}\left(3+\sqrt {15x+9}\right)\left(4x^2+13x-12\right)-2\left(3+\sqrt {15x+9}\right)=-15\end{align}
$$
$$
\begin{align}\left(3+\sqrt {15x+9}\right)\left(4x^2+13x-12\right)=2\sqrt{15x+9}-9\end{align}
$$
Then, multiplying both side of the equation by the conjugate $2\sqrt{15x+9}+9\neq 0\thinspace$, yields :
$$
\begin{align}\left(2\sqrt{15x+9}+9\right)\left(3+\sqrt{15x+9}\right)\left(x+4\right)\left(4x-3\right)=15\left(4x-3\right)\end{align}
$$
Thus, based on the equivalence between the mathematical steps, we determine that $x_2=\frac 34$ is the second real root of the original equation, since $\thinspace 4x-3\thinspace$ is the common factor of the left and right sides of the equation.
Finally, we need to solve :
$$
\begin{align}\overbrace{\left(9+2\sqrt{15x+9}\right)}^{\ge 9}\thinspace
\overbrace {\left(3+\sqrt {15x+9}\right)}^{\ge 3}\thinspace\overbrace {\left(x+4\right)}^{>3}=15\end{align}
$$
However, we see that the last equation we obtained above has no real roots. Therefore, the original equation has only $2$ real roots : $\thinspace x\in\left\{0,\frac 34\right\}\thinspace.$
This completes the solution.

$\rm {Comment:}$
Remember that, this is not correct to generalize the method we used. The method only works on specific instances. Indeed, replace $\sqrt {15x+9}$ with the radical expression $\sqrt {7x+9}$ in the original equation, then we will definitely have to apply squaring operations and use Galois theory.
A: It does make sense to square it.  You do end up with a 6th degree polynomial,
$$16x^6+104x^5+57x^4-388x^3+118x^2+69x=0$$
Then you can use the Rational Root Theorem to restrict the rational solution set to $\pm \frac{\{ 1, 3, 23, 69 \}}{\{ 1, 2, 4, 8, 16 \}}$.  It turns out that only $x = 0$ and $x = \frac{3}{4}$ are valid roots, so you can factor out $x(4x-3)$, giving:
$$4x^4 + 29 x^3 + 36 x^2 - 70 x - 23 = 0$$
A quartic equation can be solved with a formula, but I don't feel like dealing with the crazy nested radicals, so here are the numerical solutions:
$$x \approx -0.2941828832484235$$
$$x \approx 1.180434322701339$$
$$x \approx -4.068125719726457 \pm 0.09154917260014624i$$
However, only $x = 0$ and $x = \frac{3}{4}$ satisfy the original equation $4x^3+13x^2-14x=3-\sqrt{15x+9}$.  The other four roots are extraneous solutions introduced by squaring.
A: Using your hint but working with multiple angles, we end with the problem of finding the zeros of function
$$f(t)=250 \sqrt{2}\,\, | \cos (t)| -646 \cos (2 t)+195 \cos (4 t)+18 \cos (6 t)-55$$ First notice that
$$f(t)=f(\pi-t)$$ Now, the $\sqrt 2$ is quite appealing. So, we have two roots $t=\frac \pi 4$ and $t=\frac {3\pi} 4$.
For sure, what remains is to prove that there is no other solution.
