# Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$

How to solve this equition?

$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$

My attempt:

$$10x^4 - (7x^2+1)(x^2+x+1)=0$$

Thats all i can

Update

Tried to open brakets and simplify:

$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1$$ $$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4-7x^3-8x^2-1=0$$

• I recommend you expand the brackets and simplify – Obinoscopy Jul 26 '15 at 9:50
• I do not recommend that – qwr Jul 26 '15 at 9:56

$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$

Set $t=x^2,z=x^2+x+1$.

$\Longrightarrow$

\begin{align}10t^2-7tz+z^2&=(2t-z)(5t-z)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}

$$\boxed{\color{red}{x_{1,2}=\frac{1}{2}\pm\frac{\sqrt5}{2},\;x_{3,4}=\frac{1}{8}\pm \frac{\sqrt{17}}{8}}}$$

• How did you come up with this factorization? – lisyarus Jul 26 '15 at 9:59
• @mathlove has shown it – Evgeny Semyonov Jul 26 '15 at 10:09
• @Nehorai: You wrote first only $(x^2-x-1)(4x^2-x-1)=0$ without writing how you came up with this factorization. So, I wrote my answer with how. Then, you added how. What I'm saying is that "I wrote the final answer before mathlove" is a very misleading comment. – mathlove Jul 26 '15 at 10:28

Set $A=x^2,B=x^2+x+1$. Then, \begin{align}10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2&=10A^2-7AB+B^2\\&=(2A-B)(5A-B)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}

Divide by $(x^2 + x + 1)^2$, the equation becomes: $$10\frac{x^4}{(x^2 + x + 1)^2} - 7\frac{x^2}{x^2 + x + 1} + 1 = 0$$ Let $z = \frac{x^2}{x^2 + x + 1}$. The equation now is $$10z^2 - 7z + 1 = 0$$ Solving it like an ordinary quadratic equation on $z$ you get at most two roots $z_{1,2}$. Then let $\frac{x^2}{x^2 + x + 1} = z_1$, or, equivalently, $$x^2 = z_1 (x^2+x+1)$$ It is a quadratic equation in $x$. Similary for $z_2$. Solutions to this two equations are solutions to the initial problem.

• You can also directly factor $10a^2-7ab+b^2=(-5a+b)(-2a+b)$. – Yves Daoust Jul 26 '15 at 10:02
• @YvesDaoust yes, I can. mathlove has already shown this way of solution. I believe that such a factorization is not so obvious and involves either guesses or dividing and finding roots; the second way is exactly my answer. – lisyarus Jul 26 '15 at 10:06

$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$

Divide by $x^4$ on both sides,

$10-\frac{7x^2(x^2+x+1)}{x^4}+\frac{(x^2+x+1)^2}{x^4}=0$

$10-7(1+\frac{1}{x}+\frac{1}{x^2})+(1+\frac{1}{x}+\frac{1}{x^2})^2=0$

Put $(1+\frac{1}{x}+\frac{1}{x^2})=t$

$10-7t+t^2=0$,solving we get $t=2,5$

when $t=2$

$1+\frac{1}{x}+\frac{1}{x^2}=2$

simplify we get,$x^2-x-1=0$.............(1)

when $t=5$

$1+\frac{1}{x}+\frac{1}{x^2}=5$

simplify we get,$4x^2-x-1=0...........(2)$

Solve (1) and (2) and get the answer.