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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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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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