# Find the value of $x_1^6 +x_2^6$ of this quadratic equation without solving it

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.

$25x^2-5\sqrt{76}x+15=0$

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.

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$$x_1^6+x_2^6=(x_1^2+x_2^2)^3-3x_1^4x_2^2-3x_1^2x_2^4=(x_1^2+x_2^2)^3-3(x_1x_2)^2(x_1^2+x_2^2)$$ 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:

$$25(x-x_1)(x-x_2)=25x^2-25(x_1+x_2)x+25x_1x_2=25x^2-5\sqrt{76}+15$$

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

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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)

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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