Find the zeros in simplest radical form for the equation:
\[ y = \frac{1}{2}x^2 - 4 \]
Answer (3)
y= \frac{1}{2} x^2-4\\\\y=0\ \ \ \Leftrightarrow\ \ \ \frac{1}{2} x^2-4=0\ /\cdot2\ \ \ \Leftrightarrow\ \ \ x^2-8=0\\\\x^2-(2 \sqrt{2} )^2=0\ \ \ \Leftrightarrow\ \ \ (x-2 \sqrt{2} )(x+2 \sqrt{2} )=0\\\\x-2 \sqrt{2} =0\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x+2 \sqrt{2}=0\\\\x=2 \sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-2 \sqrt{2
y = 2 1 x 2 − 4 y = 0 2 1 x 2 − 4 = 0 / ⋅ 2 x 2 − 8 = 0 ( x − 8 ) ( x + 8 ) = 0 x − 8 = or x + 8 = 0 x = 8 or x = − 8 x = 4 ⋅ 2 or x = − 4 ⋅ 2 x = 2 2 or x = − 2 2
The zeros of the equation y = 2 1 x 2 − 4 are x = 2 2 and x = − 2 2 . This was found by setting the equation to zero and solving for x . The square root of 8 was simplified to get the final answers.
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