How do you evaluate (2\sqrt { 2} - \sqrt { 7} ) ^ { 2}(2√2−√7)2?

1 Answer
Grace
Dec 7, 2016

(2sqrt2 - sqrt7)^2(2√2−√7)2 simplifies to -4sqrt(14) + 15−4√14+15.

Explanation:

For the square of any binomial, it follows this general rule: (x - y)^2 = x^2 - 2xy + y^2(x−y)2=x2−2xy+y2. We can apply this here where xx is 2sqrt22√2 and yy is sqrt7√7

(2sqrt(2) - sqrt(7))^2(2√2−√7)2
= (2sqrt2)^2 - 2(2sqrt2)(sqrt7) + (sqrt7)^2=(2√2)2−2(2√2)(√7)+(√7)2
= 8 - 4sqrt(7*2) + 7=8−4√7⋅2+7
= -4sqrt(14) + 15=−4√14+15

So, our final simplified answer is -4sqrt(14) + 15−4√14+15.

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