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I got this question for homework that I have 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.

Later
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Chloe Gonzales
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3 Answers3

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\begin{align} 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). \end{align}

Since $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$, therefore: \begin{align} 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) \\ &= \left( \left(\frac{5\sqrt{76}}{25}\right)^2 -2\left(\frac{15}{25}\right) \right)^3 -3\left(\frac{15}{25}\right)^2 \left( \left(\frac{5\sqrt{76}}{25}\right)^2 -2\left(\frac{15}{25}\right) \right). \end{align}

The values of $x_1x_2$ and $x_1+x_2$ come from the following argument: \begin{align} 25(x-x_1)(x-x_2) &= 25x^2-25(x_1+x_2)x+25x_1x_2 \\ &= 25x^2-5\sqrt{76}+15. \end{align}

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

Rócherz
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Amr
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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)

Mark Bennet
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There are good methods described here and I want to add one that comes from complex analysis.

Let be $p(x)=3x^3-8x^2+2x-4$ a polynomial and $S_n=r_1^n+r_2^n+r_3^n$ the sum of its $3$ roots to the power $n$.

We can determine the values of $S_n$ by:

  • recursion as Mark Bennet answer (very efficient method)
  • iteration following a simple division.

Recursion:

If we change $x^m$ by $S_m$ in the polynomial, it will be equal to zero. it means: $$ 3S_3-8S_2+2S_1-4S_0=0 $$ but it works for all real numbers and all polynomials.

If $\ell\in\mathbb{R}$, then $ 3S_\ell-8S_{\ell-1}+2S_{\ell-2}-4S_{\ell-3}=0 $. As all knows, $S_0=$ the number of roots (polynomial's degree), $S_1=$ sum of roots, etc.

Iteration:

For this method we need to use the derivative of $p(x)$ and the Horner schema for division:

$$ \dfrac{p'(x)}{p(x)}=S_0x^{-1}+S_1x^{-2}+S_2x^{-3}+S_3x^{-4}+\dots $$

In this case, we see that $p'(x)=9x^2-16x+2$. Then:

Division by Horner of p'(x)/p(x)

          3| 9    -16    2         ---> p'(x)



p(x)-->    -8|        8   -4    12 
              2|           52/3  20/3  32/3
             -4|____________________________
               | S_0  S_1   S_2   S_3   S_4  ...

Why those methods works?

Recursive method words because $3S_3-8S_2+2S_1-4S_0= 3(r_1^3+r_2^3+r_3^3) -8(r_1^2+r_2^2+r_3^2)+2(r_1+r_2+r_3)-4(r_1^0+r_2^0+r_3^0) $

and this is obviously equal to:

$p(r_1)+p(r_2)+p(r_3)=0+0+0=0. $ The same if you multiply all by $r^{\ell-3}$. You will have the same result.

Iterative methods works because complex calculus is overpowerfull.

$p(x)=3(x-r_1)(x-r_2)(x-r_3)$ so $\ln(p(x))= \ln(3)+\ln(x-r_1)+\ln(x-r_2)+\ln(x-r_3) $

a simple derivative gives us:

$$\dfrac{p'(x)}{p(x)} = 0 + \dfrac{\color{red} 1}{x-r_1} + \dfrac{\color{red} 1}{x-r_2}+ \dfrac{\color{red} 1}{x-r_3} = \displaystyle\frac1x\sum_{1\le i\le 3} \left( \dfrac{1}{1-\frac{r_i}{x}} \right) $$ (the red one's will change for the multiplicity of the root) and $$ \displaystyle\sum_{1\le i\le 3} \left( \dfrac{1}{1-\frac{r_i}{x}} \right) = \displaystyle\sum_{1\le i\le 3} \left( 1+\frac{r_i}{x}+\frac{r_i^2}{x^2}+\frac{r_i^3}{x^3}+\cdots \right) = (1+1+1) + \frac{r_1+r_2+r_3}{x} + \frac{r_1^2+r_2^2+r_3^2}{x^2} + \cdots $$

Lutz Lehmann
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L F
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  • The last method is wrong in its first description. In the end you have clearly demonstrated that $\frac{p(x)}{p'(x)}=S_0+S_1x^{-1}+S_2x^{-2}+S_3x^{-3}+...$. So you have to divide as if the highest power of $x$ is dominant, as is apparently done in the sketched schema. Or apply the formula as given to $q(x)=x^np(1/x)$. – Lutz Lehmann Aug 16 '23 at 10:19
  • Hello @LutzLehmann . I known it works when dividing "coefficients backwards". Once I read that this has to be done like this and I'm not sure why. This trick has helped me in a lot of times for complex calculus. I cannot find a justification why we need to divide in that order. maybe because is a non-algebraic division. – L F Aug 17 '23 at 14:45
  • Sorry, the division is and remains $p'(x)/p(x)$, I was too hasty above. If you compute it for $x\approx 0$ to get a power or perturbation series for small $x$, then the coefficients you get are not the power sums for positive exponents. If you evaluate for $x\approx \infty$, then the computation follows the schema that you started, and you get the intended power sums as coefficients. – Lutz Lehmann Aug 17 '23 at 19:04
  • @LutzLehmann can you edit to improve the answer? I stopped following mathematical things since I started working on other areas – L F Aug 17 '23 at 23:50