Question

Find the zeros in simplest radical form:   y=1/2x^2-4

Answer 1

$y= \frac{1}{2} x^2-4\\\\y=0\ \ \ \Leftrightarrow\ \ \ \frac{1}{2} x^2-4=0\ /\cdot2\ \ \ \Leftrightarrow\ \ \ x^2-8=0\\\\x^2-(2 \sqrt{2} )^2=0\ \ \ \Leftrightarrow\ \ \ (x-2 \sqrt{2} )(x+2 \sqrt{2} )=0\\\\x-2 \sqrt{2} =0\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x+2 \sqrt{2}=0\\\\x=2 \sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-2 \sqrt{2$

Answer 2

$y= \frac{1}{2}x^2-4\\ \\ y =0 \\ \\\frac{1}{2}x^2-4 =0 \ \ / \cdot 2\\ \\x^2-8=0 \\ \\(x-\sqrt{8})(x+\sqrt{8})=0 \\ \\ x-\sqrt{8}= \ \ or \ \ x+\sqrt{8} = 0 \\ \\x=\sqrt{8} \ \ or \ \ x=-\sqrt{8} \\ \\x=\sqrt{4\cdot 2} \ \ or \ \ x= -\sqrt{4\cdot 2}\\ \\ x=2\sqrt{2} \ \ or \ \ x=-2\sqrt{2}$