5
$\begingroup$

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

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.

$\endgroup$
  • 6
    $\begingroup$ Let $y=x^2 - 2x + 3$. Then $y^2 = 121$. Solve for $y$, then solve for $x$. $\endgroup$ – Mark Sep 21 '15 at 0:59
  • 1
    $\begingroup$ @Mark why don't you expand that into a full answer? $\endgroup$ – davidlowryduda Sep 21 '15 at 1:02
  • $\begingroup$ Yes Mark is right , solve it and you will get the answers: -2 , 4, 1 - 13^(1/2)*1i, 1 + 13^(1/2)*1i $\endgroup$ – Syed Alam Abbas Sep 21 '15 at 1:03
  • 4
    $\begingroup$ The equation you were given has a very nice structure. Multiplying out, beside being a lot of work, destroys structure. $\endgroup$ – André Nicolas Sep 21 '15 at 3:01
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You have a typo in your solution, you wrote down $y_0^2 = 122$ instead of $y_0^2 = 121$. $\endgroup$ – Aldon Sep 21 '15 at 1:20
  • $\begingroup$ @Aldon Of course, thanks for pointing that out $\endgroup$ – Mark Sep 21 '15 at 1:21
6
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

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.