-3

$$(x^2+x+1)^2=x^2(3x^2+x+1)$$

In addition to using ai solvers are able to give me the final solution for this equation. But I am not able to get the exact steps or procedure to reach the solution. I need a clear step by step procedure to attain the solution. So I seek this community with esteemed mathematicians to solve my problem

Ryka
  • 23
  • 1
    Welcome to MSE. Please update your question text to show what you tried (in addition to using "ai solvers"), and especially anything you had difficulty with, to determine a final solution yourself. Thanks. – John Omielan Aug 09 '19 at 03:55
  • Where does this problem come from? Please see how to ask a good question. – Toby Mak Aug 09 '19 at 03:57
  • @TobyMak bro the problem is I don't know how did the answer come so you please help if you can by solve the question step by step please post it – Ryka Aug 09 '19 at 03:59
  • 1
    Sorry, but most people on this site won't answer unless you have sufficient context. You are welcome to post this question on another site. – Toby Mak Aug 09 '19 at 04:00
  • I don't need most of the people to answer. I only need the person who are ready help the beginners. Every one cannot be a pro in the beginning – Ryka Aug 09 '19 at 04:02
  • @TobyMak Do I count as an exception? – Parcly Taxel Aug 09 '19 at 04:04
  • I am only judging on the quality of your answer, since it does not give too much to the OP. When this question is closed and possibly deleted, you will lose the upvote anyway. – Toby Mak Aug 09 '19 at 04:06
  • @Ryka Yes, I understand. In fact Math SE is for all levels of maths, as written in the tour. However, your question is not a good question by our site's standards, so you can either improve it, or ask elsewhere. – Toby Mak Aug 09 '19 at 04:09

2 Answers2

3

We can rewrite the equation as $$2x^4-x^3-2x^2-2x-1=0$$ which can be factored into two quadratics (by any of several hand methods): $$(x^2-x-1)(2x^2+x+1)=0$$ Solving these quadratics yields the roots $\frac{1\pm\sqrt5}2$ and $\frac{-1\pm\sqrt7i}4$.

Parcly Taxel
  • 105,904
0

It's $$(x^2+x+1)^2=x^2(x^2+x+1)+2x^4$$ or $$(x^2+x+1)^2-x^2(x^2+x+1)-2x^4=0,$$ which gives $$x^2+x+1=-x^2,$$ which is impossible for real $x$, or $$x^2+x+1=2x^2.$$ Can you end it now?

The last step we cam make by the following way.

Let $x^2+x+1=a$.

Thus, we need to sove $$a^2-x^2a-2x^4=0$$ or $$a^2-2x^2a+x^2a-2x^4=0$$ or $$a(a-2x^2)+x^2(a-2x^2)=0$$ or $$(a+x^2)(a-2x^2)=0,$$ which gives $$a=-x^2$$ or $$a=2x^2.$$

Michael Rozenberg
  • 203,855