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

• chloe1995
In summary, the conversation discusses solving for x_1^6+x_2^6 for a quadratic equation without actually solving for the equation. The individual mentions trying to factor the equation but only rewrites it and asks for guidance on how to proceed without solving the equation. The response suggests using the sum and product of roots formula to find the value of x_1 + x_2 and x_1\cdot x_2.
chloe1995

## Homework Statement

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.

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

## 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.

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.

chloe1995 said:

## Homework Statement

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.

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

## 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}\ .\$

SteamKing said:
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.

SammyS said:
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.

So, Have you managed to solve the problem?

## What is the purpose of finding the value of x1^6 + x2^6 without solving the quadratic equation?

The purpose of finding the value of x1^6 + x2^6 without solving the quadratic equation is to determine the sum of the sixth powers of the roots of the equation. This value can provide insight into the behavior and characteristics of the roots, and can be useful in solving other problems involving the equation.

## Can the value of x1^6 + x2^6 be found without solving the quadratic equation?

Yes, it is possible to find the value of x1^6 + x2^6 without solving the quadratic equation. This can be done using the Vieta's formulas, which relate the coefficients of a polynomial to the sums and products of its roots.

## What are the Vieta's formulas and how are they used to find the value of x1^6 + x2^6?

The Vieta's formulas state that for a polynomial of degree n, the sum of the roots is equal to the negative of the coefficient of the (n-1)th term, and the product of the roots is equal to the constant term. Therefore, to find the value of x1^6 + x2^6, we can use the sums and products of the roots of the quadratic equation to calculate the sixth powers and then add them together.

## Are there any limitations to using the Vieta's formulas to find the value of x1^6 + x2^6?

Yes, there are limitations to using the Vieta's formulas. They can only be used for polynomials with real coefficients, and they may not provide an exact solution due to rounding errors. Additionally, the Vieta's formulas only work for quadratic equations with two distinct roots.

## What other applications can the value of x1^6 + x2^6 be used for?

The value of x1^6 + x2^6 can be used in problems involving the manipulation of polynomial equations, such as finding the roots of a polynomial or determining the coefficients of a polynomial given its roots. It can also be used in fields such as physics and engineering to solve problems involving quadratic equations.

• Precalculus Mathematics Homework Help
Replies
22
Views
2K
• Precalculus Mathematics Homework Help
Replies
27
Views
3K
• Precalculus Mathematics Homework Help
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
32
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
647
• Precalculus Mathematics Homework Help
Replies
2
Views
1K
• Precalculus Mathematics Homework Help
Replies
1
Views
890
• Precalculus Mathematics Homework Help
Replies
7
Views
1K
• Precalculus Mathematics Homework Help
Replies
14
Views
2K
• Precalculus Mathematics Homework Help
Replies
12
Views
2K