Question
What are the roots of the polynomial function h(x)=3x^2+243 ? (1 point) The roots are 9 and -9. The roots are 81 and -81. The roots are 9i and -9i. The roots are 81i and -81i.
Answers
Answered by
GPT-4o mini
To find the roots of the polynomial function \( h(x) = 3x^2 + 243 \), we need to set the equation equal to zero:
\[
3x^2 + 243 = 0
\]
First, we can isolate \( x^2 \):
\[
3x^2 = -243
\]
Now, divide both sides by 3:
\[
x^2 = -81
\]
To find \( x \), take the square root of both sides:
\[
x = \pm \sqrt{-81} = \pm \sqrt{81} \cdot \sqrt{-1}
\]
Since \( \sqrt{81} = 9 \) and \( \sqrt{-1} = i \), we get:
\[
x = \pm 9i
\]
Thus, the roots of the polynomial function \( h(x) = 3x^2 + 243 \) are:
**The roots are \( 9i \) and \( -9i \).**
\[
3x^2 + 243 = 0
\]
First, we can isolate \( x^2 \):
\[
3x^2 = -243
\]
Now, divide both sides by 3:
\[
x^2 = -81
\]
To find \( x \), take the square root of both sides:
\[
x = \pm \sqrt{-81} = \pm \sqrt{81} \cdot \sqrt{-1}
\]
Since \( \sqrt{81} = 9 \) and \( \sqrt{-1} = i \), we get:
\[
x = \pm 9i
\]
Thus, the roots of the polynomial function \( h(x) = 3x^2 + 243 \) are:
**The roots are \( 9i \) and \( -9i \).**