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3 years ago

9

What are the zeros of h(x)=x^4-18x^2+81

2 answers:

kykrilka [37]3 years ago

5 0

Answer:

zeros: x = -3, 3

Step-by-step explanation:

The zeros are the x values where the graph intersects the x-axis. to find the roots, replace the y with 0 and solve for x.

Mark Brainliest?

<h2>HOPE THIS HELPS!</h2>

vitfil [10]3 years ago

3 0

$Answer: \\ Let \: t \: be \: {x}^{2} \\The \: equation \: is: {t}^{2} - 18t + 81 = 0 (to \: find \: the \: zeros \: we \: set \: y = 0)\\ \Leftrightarrow \: {(t - 9)}^{2} = 0 \Leftrightarrow t - 9 = 0 \Leftrightarrow t = {x}^{2} = 9 \\ \Leftrightarrow x = 3 \: \vee \: x = - 3 \\ \Rightarrow \: Zeros: - 3,3$

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