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
Best Answer
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