kobiquinn11 kobiquinn11

20-05-2014
Mathematics

Answered

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

Please help!

Verified Answer :

20-05-2014

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

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kate200468 kate200468

21-05-2014

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

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