# Find the value of x1^6 +x2^6 of this quadratic equation without solving it

by chloe1995
 P: 2 1. The problem statement, all variables and given/known data 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. $25x^2-5\sqrt{76}x+15=0$ 2. Relevant equations 3. The attempt at a solution 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.
 HW Helper Thanks P: 4,274 Notice that 5 can be factored from the quadratic without changing the roots. Also, you haven't truly factored the quadratic, you have merely re-written it.
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 Quote by chloe1995 1. The problem statement, all variables and given/known data 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. $25x^2-5\sqrt{76}x+15=0$ 2. Relevant equations 3. The attempt at a solution 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.
Hello chloe1995. Welcome to PF !

Suppose that x1 and x2 are the solutions to the quadratic equation, $\displaystyle \ \ ax^2+bx+c=0\ .$

Then $\displaystyle \ \ x_1 + x_2 = -\frac{b}{a}\ \$ and $\displaystyle \ \ x_1\cdot x_2=\frac{c}{a}\ .\$

P: 2

## Find the value of x1^6 +x2^6 of this quadratic equation without solving it

 Quote by SteamKing Notice that 5 can be factored from the quadratic without changing the roots. Also, you haven't truly factored the quadratic, you have merely re-written it.
Oops! I meant completing the square.

 Quote by SammyS Hello chloe1995. Welcome to PF ! Suppose that x1 and x2 are the solutions to the quadratic equation, $\displaystyle \ \ ax^2+bx+c=0\ .$ Then $\displaystyle \ \ x_1 + x_2 = -\frac{b}{a}\ \$ and $\displaystyle \ \ x_1\cdot x_2=\frac{c}{a}\ .\$
Thank you.
 PF Patron HW Helper Sci Advisor Emeritus P: 7,077 So, Have you managed to solve the problem?

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