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

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

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Answer 1 · 2016-12-07T03:23:21

${(2 \sqrt{2} - \sqrt{7})}^{2}$ $(2sqrt2 - sqrt7)^2$ simplifies to $- 4 \sqrt{14} + 15$ $-4sqrt(14) + 15$ .

Explanation:

For the square of any binomial, it follows this general rule: ${(x - y)}^{2} = x^{2} - 2 x y + y^{2}$ $(x - y)^2 = x^2 - 2xy + y^2$ . We can apply this here where $x$ $x$ is $2 \sqrt{2}$ $2sqrt2$ and $y$ $y$ is $\sqrt{7}$ $sqrt7$

${(2 \sqrt{2} - \sqrt{7})}^{2}$ $(2sqrt(2) - sqrt(7))^2$
$= {(2 \sqrt{2})}^{2} - 2 (2 \sqrt{2}) (\sqrt{7}) + {(\sqrt{7})}^{2}$ $= (2sqrt2)^2 - 2(2sqrt2)(sqrt7) + (sqrt7)^2$
$= 8 - 4 \sqrt{7 \cdot 2} + 7$ $= 8 - 4sqrt(7*2) + 7$
$= - 4 \sqrt{14} + 15$ $= -4sqrt(14) + 15$

So, our final simplified answer is $- 4 \sqrt{14} + 15$ $-4sqrt(14) + 15$ .

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

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