All the roots of $(x^2+1)^2 = x(3x^2+4x+3)$

Question

Find all the roots of the equation : $$(x^2+1)^2 = x(3x^2+4x+3)$$How do we find the roots in polynomials of degree > 2 ??

Also, In odd degree polynomials I use Descartes rule of signs to predict the number of real roots. Here, however, it does not give me any clue about the number of real/imaginary roots. Is there a better method?

score 7 · Accepted Answer · edited Apr 13 '17 at 12:20

7

On rearrangement we have $x^4-3x^3-2x^2-3x+1=0$

Like this, divide either sides by $x^2$ as $x\ne0$ to get $$x^2+\frac1{x^2}-3\left(x+\frac1x\right)-2=0$$

Or, $$\left(x+\frac1x\right)^2-2-3\left(x+\frac1x\right)-2=0$$

Put $x+\frac1x=u$

edited Apr 13 '17 at 12:20

Community

1

answered Sep 01 '13 at 07:59

lab bhattacharjee

274,582

How can we assume $x\ne0$ – Simar Sep 01 '13 at 08:01
@Simar, put $x=0$ in the given eqaution – lab bhattacharjee Sep 01 '13 at 08:02
k. This gives me $x^2 - 3x - 2- 3/x + 1/x^2$ . Further? – Simar Sep 01 '13 at 08:03
@Simar, please find the edited answer – lab bhattacharjee Sep 01 '13 at 08:04
Ah! Got that. What about the number of real roots part ?(In general for higher degree polynomials?) – Simar Sep 01 '13 at 08:10
@Simar, observe that for real $x,$ if $u=x+\frac1x\iff x^2-xu+1=0\implies u^2\ge4\cdot1\cdot1=4\implies u\le-2$ or $u\ge2$ .Here observe that $u=4,-1$ – lab bhattacharjee Sep 01 '13 at 08:13

score 7 · Answer 2 · answered Sep 01 '13 at 08:04

7

Hint: $$(x^2+1)^2-3x(x^2+1)-4x^2=0$$ $$(x^2+1-4x)(x^2+1+x)=0$$

answered Sep 01 '13 at 08:04

njguliyev

14,473
1
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Great. Please help me on the second part of the question. – Simar Sep 01 '13 at 08:09

score 3 · Answer 3 · answered Sep 01 '13 at 08:07

3

Another solution:

Solve the equation $(x^2+1)^2 = 3x(x^2+1)+4x^2$ with $x^2+1$ as an unknown: $$ x^2+1=\frac{3x\pm \sqrt{9x^2+16x^2}}{2}=\frac{3x\pm 5x}{2} $$ etc.

answered Sep 01 '13 at 08:07

Boris Novikov

17,470

All the roots of $(x^2+1)^2 = x(3x^2+4x+3)$

3 Answers3