How would I solve this higher degree equation?

Question

$$(x^{ 2 }-2x+3)(x^{ 2 }-2x+3)=121$$

Steps I took:

$$x^4-2x^3+3x^2-2x^3+4x^2-6x+3x^2-6x+9=121$$

$$x^{ 4 }-4x^{ 3 }+10x^{ 2 }-12x+9=121$$

$$x^{ 4 }-4x^{ 3 }+10x^{ 2 }-12x-112=0$$

At this point I don't see how I can possibly find the solution using either the quadratic formula, factoring, or completing the square. Please do not provide the solution. I only want hints to guide me in the right direction.

Let $y=x^2 - 2x + 3$. Then $y^2 = 121$. Solve for $y$, then solve for $x$. — Mark, Sep 21 '15 at 00:59
Yes Mark is right , solve it and you will get the answers: -2 , 4, 1 - 13^(1/2)1i, 1 + 13^(1/2)1i — Syed Alam Abbas, Sep 21 '15 at 01:03
The equation you were given has a very nice structure. Multiplying out, beside being a lot of work, destroys structure. — André Nicolas, Sep 21 '15 at 03:01

score 6 · Answer 1 · answered Sep 21 '15 at 01:03

Outline :

No need to complicate matters. Observe both LHS and RHS are perfect squares.

So either $(x^2 -2x + 3) = 11$ or $(x^2 -2x + 3) = -11$ which means now you are solving $x^2 -2x - 8 = 0$ or $x^2 -2x + 14 = 0$ which are 2 quadratics, which you can solve

Mark · Accepted Answer · 2015-09-21T01:21:02.723

5

Let $y = x^2 - 2x + 3$. Then our original equation is $y^2 = 121$. Solving this equation gives us the solutions $\pm y_0$, where $y^2_0 = 121$. Then solve the two equations $y = y_0$ for $x$. In the end, you'll get 4 solutions that satisfy your original equation.

edited Sep 21 '15 at 01:21

answered Sep 21 '15 at 01:07

Mark

2,535

You have a typo in your solution, you wrote down $y_0^2 = 122$ instead of $y_0^2 = 121$. – mopy Sep 21 '15 at 01:20
@Aldon Of course, thanks for pointing that out – Mark Sep 21 '15 at 01:21

score 2 · Answer 3 · answered Sep 21 '15 at 06:52

Note that $(x^2 - 2x + 3)(x^2 - 2x + 3) = (x^2 - 2x + 3)^2$. Writing it this way will allow you to square-root both sides and then apply your other procedures.

As a lesson, don't just blindly simplify everything as you tried to do; be on the lookout for opportunities to write things as squares (frequently binomial squares), which is really the core idea behind the method of square roots and completing the square.

How would I solve this higher degree equation?

3 Answers3