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How to solve the following equation?

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


My attempt:

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

Thats all i can

Update

Tried to open brakets and simplify:

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

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  • $\begingroup$ I recommend you expand the brackets and simplify $\endgroup$
    – Obinoscopy
    Jul 26, 2015 at 9:50
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    $\begingroup$ I do not recommend that $\endgroup$
    – qwr
    Jul 26, 2015 at 9:56

4 Answers 4

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

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

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

$\Longrightarrow$

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

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

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    $\begingroup$ How did you come up with this factorization? $\endgroup$
    – lisyarus
    Jul 26, 2015 at 9:59
  • $\begingroup$ @mathlove has shown it $\endgroup$ Jul 26, 2015 at 10:09
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    $\begingroup$ @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. $\endgroup$
    – mathlove
    Jul 26, 2015 at 10:28
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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.

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  • $\begingroup$ You can also directly factor $10a^2-7ab+b^2=(-5a+b)(-2a+b)$. $\endgroup$
    – user65203
    Jul 26, 2015 at 10:02
  • $\begingroup$ @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. $\endgroup$
    – lisyarus
    Jul 26, 2015 at 10:06
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$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$

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.

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