4
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ I recommend you expand the brackets and simplify $\endgroup$ – Obinoscopy Jul 26 '15 at 9:50
  • 6
    $\begingroup$ I do not recommend that $\endgroup$ – qwr Jul 26 '15 at 9:56
8
$\begingroup$

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

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

$\Longrightarrow$

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

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

$\endgroup$
  • 2
    $\begingroup$ How did you come up with this factorization? $\endgroup$ – lisyarus Jul 26 '15 at 9:59
  • $\begingroup$ @mathlove has shown it $\endgroup$ – Evgeny Semyonov Jul 26 '15 at 10:09
  • 3
    $\begingroup$ @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 '15 at 10:28
9
$\begingroup$

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

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You can also directly factor $10a^2-7ab+b^2=(-5a+b)(-2a+b)$. $\endgroup$ – Yves Daoust Jul 26 '15 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 '15 at 10:06
5
$\begingroup$

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.