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$99\sqrt{2}$ is a root of $f(x) = x^2 - 19,602$ . 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.

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Conjugate root theorems

Full solution

Q.

99\sqrt{2}

is a root of

f(x) = x^2 - 19,602

. 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.

Roots Analysis: Since $99\sqrt{2}$ is a root, the other root will also be a real number because the coefficients of the polynomial are real.
Sum of Roots: The sum of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $-\frac{b}{a}$ . For $f(x) = x^2 - 19,602$ , $a = 1$ and $b = 0$ , so the sum of the roots is $0$ .
Finding Second Root: If one root is $99\sqrt{2}$ , the other root must be $-99\sqrt{2}$ to sum to $0$ .
Product of Roots Check: Check the product of the roots: $99\sqrt{2}$ * $-99\sqrt{2}$ = $-19$ , $602$ . This matches the constant term $c$ of the polynomial, confirming our root is correct.

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