77sqrt 3 is a root of f(x) = x^2 - 17,787. Find the other roots of f(x). Write your answer as a list of simplified values separated by commas, if there is more than one value. ______

Apply Conjugate Root Theorem: Since 77sqrt(3) is a root, we can use the conjugate root theorem which states that if a polynomial has real coefficients and a non-real root, then its conjugate is also a root.Identify Conjugate Root: The conjugate of 77sqrt(3) is 77sqrt(3) with a negative sign, so the other root is -77sqrt(3).List Polynomial Roots: Now we list the roots of the polynomial: 77sqrt(3), -77sqrt(3).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$77\sqrt{3}$ is a root of $f(x) = x^2 - 17,787$ . Find the other roots of $f(x)$ . $\newline$ Write your answer as a list of simplified values separated by commas, if there is more than one value. $\newline$ ______ $\newline$

View step-by-step help

Home

Math Problems

Precalculus

Conjugate root theorems

Full solution

77\sqrt{3}

is a root of

f(x) = x^2 - 17,787

. Find the other roots of

f(x)

\newline

Write your answer as a list of simplified values separated by commas, if there is more than one value.

\newline

______

\newline

Apply Conjugate Root Theorem: Since $77\sqrt{3}$ is a root, we can use the conjugate root theorem which states that if a polynomial has real coefficients and a non-real root, then its conjugate is also a root.
Identify Conjugate Root: The conjugate of $77\sqrt{3}$ is $77\sqrt{3}$ with a negative sign, so the other root is $-77\sqrt{3}$ .
List Polynomial Roots: Now we list the roots of the polynomial: $77\sqrt{3}$ , $-77\sqrt{3}$ .