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Find the zeros of the function. h(x)=2x²+72 The zeros of h are x=

User Harnex
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Explanation:

To find the zeros of the function \(h(x) = 2x^2 + 72\), set the function equal to zero and solve for \(x\):

\[2x^2 + 72 = 0\]

First, divide both sides of the equation by 2 to simplify it:

\[x^2 + 36 = 0\]

Now, subtract 36 from both sides:

\[x^2 = -36\]

Since the right side is negative, this equation has no real solutions because the square of any real number is always non-negative. However, if you allow for complex numbers, you can take the square root of both sides:

\[x = \pm 6i\]

So, the zeros of the function \(h(x) = 2x^2 + 72\) are \(x = 6i\) and \(x = -6i\), where \(i\) is the imaginary unit.

User Ong
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