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.

Solution 1:

\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 $x_1+x_2$ .

Solution 2:

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)