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Solve for x: (x – 4)2 = 2
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r u squaring that or multiplying?
\[(x-4)^2 =2\] is this your question?
(x-4)2=2 2x-8=2 2x=8+2 2x=10 \[2x \div2=10\div2\] x=5
dat is if your question is multiplying 2
it is ^2
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then its: (x-4)(x+4)=2 x(x+4)-4(x+4)=2 \[x ^{2}+4x-4x-16=2\] \[x ^{2}-16=2\] \[x ^{2}=16+2\] \[x ^{2}=18\] now you square both sides and the square root will cancel the ^2 on x \[x=\sqrt{18}\] x=4.24
or \[3\sqrt{2}\] its the same but the other answer was in decimals
If that's an exponent, \((x-4)^2=2\), then \(x-4=\pm\sqrt{2}\), so \(x=4\pm\sqrt{2}\).
no you solve the set that is (x-4)^2 as different of 2 squares,then you equate it to 2
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