How do you solve 2x^2-32=0 using square roots? Please help!

asked Apr 10, 2015 66.7k views

4 votes

How do you solve 2x^2-32=0 using square roots?

Please help!

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4 votes

$2x^2-32=0\\\\2(x^2-16)=0\ /:x^2-16=0\\\\x^2-4^2=0\\\\(x-4)(x+4)=0\\\\x-4=0\ \ \ \ or\ \ \ \ \ x+4=0\\\\x=4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-4$

Kamillpg answered Apr 11, 2015

8.2k points

2 votes

$2x^2-32=0 \ \ / :2 \\ \\ x^2-16=0 \\ \\x^2-4^2 = 0 \\ \\ (x-4)(x+4)=0 \\ \\ x-4 = 0 \ \ or \ \ x+4 = 0\\ \\ x=4\ \ or \ \ x=-4$

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