Solve equation $(3x^2-2x-1)^2-(6x^2-4x-5)^2+x=0$

Question

I tried to expand and I had

$27x^4-36x^3-42x^2+35x+24=0$

I don't know how to solve this equation, but I sketch on graph and I have four irrational roots. Which method I need to use in this equation??

$x_1= \frac{\sqrt{41}}{6}+\frac{1}{2}$

$x_2= -\frac{\sqrt{41}}{6}+\frac{1}{2}$

$x_3= \frac{1-\sqrt{37}}{6}$

$x_4= \frac{1+\sqrt{37}}{6}$

This is roots of equaition, which I tried to find

Answer 1

First, let's denote $a =3x^2-2x-1$, then $$\begin{align} &\iff a^2-(2a-3)^2+x =0\\ &\iff a^2 - \left(4a^2-12a+9 \right) + x = 0 \\ &\iff -3a^2+12a-9 + x = 0 \\ &\iff -9a^2+36a-27 + 3x = 0 \\ &\iff -(3a)^2+2\cdot3a\cdot6-6^2+9 + 3x = 0 \\ &\iff -(3a-6)^2+ 3x +9= 0 \tag{1} \end{align}$$ Replace $a =3x^2-2x-1$, then: $$\begin{align} (1)&\iff -(3(3x^2-2x-1)-6)^2+ 3x +9= 0 \\ &\iff -(9x^2-6x-9)^2+ 3x +9= 0 \\ &\iff -((3x-1)^2-10)^2+ (3x-1) +10= 0 \tag{2} \end{align}$$ Denote $b=3x-1$, then: $$\begin{align} (2)&\iff (b^2-10)^2- (b +10)= 0 \\ &\iff (b^2-10-b)(b^2+b-9)=0 \tag{3} \end{align}$$ From $(3)$, we deduce that the initial equations has $4$ roots which are roots of the two quadratic equations below: $$\begin{align} &(A):b^2-10-b=0 \iff (3x-1)^2-10-(3x-1)=0 \iff \color{red}{x\in\left\{\frac{1}{2}\pm\frac{\sqrt{41}}{6} \right\}}\\ &(B):b^2-9+b=0 \iff (3x-1)^2-9+(3x-1)=0 \iff \color{red}{x\in\left\{\frac{1}{6}\pm\frac{\sqrt{37}}{6} \right\}} \end{align}$$

Q.E.D

Answer 2

Let\begin{align}p(x)&=x^4-\frac43x^3-\frac{14}9x^2+\frac{35}{27}x+\frac89\\&=\frac1{27}\left(27x^4-36x^3-42x^2+35x+24\right).\end{align}Then the reduced form of $p(x)$ is\begin{align}q(x)&=p\left(x+\frac13\right)\\&=y^4-\frac{20}9y^2-\frac1{27}y+\frac{10}9.\end{align}Let $a_0=\frac{10}9$, $a_1=-\frac1{27}$, and $a_2=-\frac{20}9$. Then the resolvent cubic of $q(x)$ is$$r(x)=x^3+2a_2x^2+(a_2^{\,2}-4a_0)y-a_1^{\,2}=x^3-\frac{40}9x^2+\frac{40}{81}x-\frac1{729}.\label{a}\tag1$$If there is a rational number $\alpha$ such that $\alpha^2$ is a root of \eqref{a}, and if you define$$\beta=\frac12\left(a_2+\alpha^2-\frac{a_1}\alpha\right)\quad\text{and}\quad\gamma=\frac12\left(a_2+\alpha^2+\frac{a_1}\alpha\right),$$then $q(x)=(x^2+\alpha x+\beta)(x^2-\alpha x+\gamma)$. And it turns out that, yes, \eqref{a} has a root which is the square of a rational, since $\frac19$ is a root of \eqref{a} (which can be guessed from the rational root theorem). So, $\alpha=\frac13$, $\beta=-1$, and $\gamma=-\frac{10}9$. Therefore$$q(x)=\left(x^2+\frac13x-1\right)\left(x^2-\frac13x-\frac{10}9\right)$$and all that remains to be done is to compute the roots of these two quadratics and to use the fact that$$p(x)=0\iff q\left(x-\frac13\right)=0.$$

Answer 3

After expanding we get: $$27x^4-36x^3-42x^2+35x+24=0$$ If $t=3x$ then $$\underbrace{t^4-4t^3-14t^2+35t+72}_{p(t)}=0$$ Now we try to find real $a,b,c,d$ such that $$p(t) = (t^2+at+b)(t^2+ct+d) $$ Now we have to solve a following system \begin{align} a+c &= -4\\ ac +b+d &=-14\\ ad+bc &= 35\\ bd&=72 \end{align} If trying to find integer coefficients we get after some trial and error $b=-8$, $d=-9$, $a=-3$ and $c=-1$. The rest is easy.

Answer 4

I would want to share a solution using the Ferrari's method.

We have the following biquadratic equation:

$$27x^4-36x^3=42x^2-35x-24$$

We attempt to make the LHS a perfect square by adding an $x^2$ term on both sides.

$$(3\sqrt 3x^2)^2-2(3\sqrt3x^2)(2\sqrt 3x)+(2\sqrt3x)^2\\ =54x^2-35x-24$$

Now that we have completed the square on LHS, we have:

$$3(3 x^2-2x)^2=54x^2-35x-24$$

Now we attempt to complete the squares on RHS as well by introducing $\lambda$.

$$3(3 x^2-2x+\lambda)^2\\=54x^2-35x-24+3\{\lambda^2+2\lambda(3x^2-2)\}\\ =\underbrace{(54+18\lambda)}_{=\ a}x^2+\underbrace{(-35-12\lambda)}_{=\ b}x+\underbrace{(3\lambda^2-24)}_{=\ c}$$

Here's how we make the RHS a perfect square:

We set its discriminant i.e., $b^2-4ac$ to be zero. $$(35+12\lambda)^2-4(54+18\lambda)(3\lambda^2-24)=0$$ This is a cubic equation. $$216\lambda^3+504\lambda^2-2568\lambda-6409=0$$ You could use Cardano's method or apply the rational root theorem to find that $\lambda=-17/6$ is a real solution.
The RHS of the original equation now becomes $a\left(x-\frac{b}{2a}\right)^2$.

$$\not 3\left(3x^2-2x-\frac{17}{6}\right)^2=\not 3\left(x-\frac{1}{6}\right)^2$$

With difference of squares, you get two quadratics.

$$(3x^2-x-3)\left(3x^2-3x-\frac{16}{6}\right)=0$$

We apply the quadratic formula to solve each of them to get the four roots of our original equation.