Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I got this question for homework and I've never seen anything similar to it.

Solve for $x_1^6+x_2^6$ for the following quadratic equation where $x_1$ and $x_2$ are the two real roots and $x_1 > x_2$, without solving the equation.


I tried factoring it and I got $(-5x+\sqrt{19})^2-4=0$

What can I do afterwards that does not constitute as solving the equation? Thanks.

share|cite|improve this question
up vote 6 down vote accepted

$$x_1^6+x_2^6=(x_1^2+x_2^2)^3-3x_1^4x_2^2-3x_1^2x_2^4=(x_1^2+x_2^2)^3-3(x_1x_2)^2(x_1^2+x_2^2)$$ Since $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$, therefore: $$x_1^6+x_2^6=((x_1+x_2)^2-2x_1x_2)^3-3(x_1x_2)^2((x_1+x_2)^2-2x_1x_2)$$ $$x_1^6+x_2^6=((\frac{5\sqrt{76}}{25})^2-2(\frac{15}{25}))^3-3(\frac{15}{25})^2((\frac{5\sqrt{76}}{25})^2-2(\frac{15}{25}))$$

The values of $x_1x_2,x_1+x_2$ come from the following argument:


Now equate the cofficents of both polynomials to get the values of $x_1x_2,x_1+x_2$

share|cite|improve this answer
I am not sure if this is the fastest way to do it. – Amr Jan 23 '13 at 19:31
It would probably be helpful to show from where the values for $x_1 x_2$ and $x_1^2 + x_2^2$ come. – JavaMan Jan 23 '13 at 19:31
OK I will include this – Amr Jan 23 '13 at 19:32
You're using a quadratic different from the OP's. – Math Gems Jan 23 '13 at 19:57
@Math Gems I edited it. – Amr Jan 23 '13 at 19:59

If you let $P_n=x_1^n+x_2^n$ then you get (multiply the equation through by $x^{n-2}$ and substitute)

$$25x_1^n-5\sqrt{76}x_1^{n-1}+15x_1^{n-2}=0$$ $$25x_2^n-5\sqrt{76}x_2^{n-1}+15x_2^{n-2}=0$$

Now add the two to get the recurrence:$$25P_n-5\sqrt{76}P_{n-1}+15P_{n-2}=0$$

$x_1+x_2$ can be read off from the equation.

$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$ or you can use $P_0=2$ to start the recurrence.

I don't suggest this as the most efficient way of solving this particular problem - but it is sometimes good to know.

ADDED in EDIT in response to comment

To add the $x_1$ and $x_2$ expressions we have $$25x_1^n-5\sqrt{76}x_1^{n-1}+15x_1^{n-2}+25x_2^n-5\sqrt{76}x_2^{n-1}+15x_2^{n-2}$$$$ =25(x_1^n+x_2^n)-5\sqrt{76}(x_1^{n-1}+x_2^{n-1})+15(x_1^{n-2}+x_2^{n-2})$$$$=25P_n-5\sqrt{76}P_{n-1}+15P_{n-2}$$

Note also that $P_0=x_1^0+x_2^0=1+1=2$ (if the constant term of the polynomial were 0, you'd have zero as a root, which would never contribute anything to the sum, so you divide through by the smallest power of $x$ to give a non-zero constant term, and proceed with a polynomial of lower degree)

share|cite|improve this answer
nice answer. As it shows what one should do with polynomials of higher degrees not just quadratic polynomials (+1) – Amr Jan 23 '13 at 19:49
This seems very interesting but I don't fully understand it. I lost you at "add the two to get the recurrence". What do you mean by recurrence and what is $P_0$? Definitely seems like a neat trick to know. Thanks. – Chloe Gonzales Jan 24 '13 at 2:49
I've added some pieces at the end of the answer. – Mark Bennet Jan 24 '13 at 7:21
Thank you. Much more clear now. – Chloe Gonzales Jan 30 '13 at 2:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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