How do you simplify $(\sqrt{x} - \sqrt{7}) (\sqrt{x} + \sqrt{7})$ $(sqrtx-sqrt7)(sqrtx+sqrt7)$ ?

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How do you simplify $(\sqrt{x} - \sqrt{7}) (\sqrt{x} + \sqrt{7})$ $(sqrtx-sqrt7)(sqrtx+sqrt7)$ ?

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Answer 1 · 2016-09-05T17:17:08

$x - 7$ $x - 7$

Explanation:

We have: $(\sqrt{x} - \sqrt{7}) (\sqrt{x} + \sqrt{7})$ $(sqrt(x) - sqrt(7)) (sqrt(x) + sqrt(7))$

Let's expand the parentheses:

$= (\sqrt{x}) (\sqrt{x}) + (\sqrt{x}) (\sqrt{7}) + (- \sqrt{7}) (\sqrt{x}) + (- \sqrt{7}) (\sqrt{7})$ $= (sqrt(x)) (sqrt(x)) + (sqrt(x)) (sqrt(7)) + (- sqrt(7)) (sqrt(x)) + (- sqrt(7)) (sqrt(7))$

$= x + \sqrt{7 x} - \sqrt{7 x} - 7$ $= x + sqrt(7 x) - sqrt(7 x) - 7$

$= x - 7$ $= x - 7$

Answer 2 · 2016-09-05T18:39:21

$x - 7$ $x-7$

Explanation:

Recall: difference of squares can be factored:

$x^{2} - y^{2} = (x + y) (x - y)$ $x^2 - y^2 = (x+y)(x-y)$

The reverse is also true, if 2 brackets are the same with different signs, we find that:

$(m + n) (m - n) = m^{2} - n^{2}$ $(m+n)(m-n) = m^2 -n^2$

This is what we have in this question,

$(\sqrt{x} - \sqrt{7}) (\sqrt{x} + \sqrt{7}) = {(\sqrt{x})}^{2} - {(\sqrt{7})}^{2}$ $(sqrtx -sqrt7)(sqrtx + sqrt7) = (sqrtx)^2 - (sqrt7)^2$

= $x - 7$ $x-7$

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How do you simplify $(\sqrt{x} - \sqrt{7}) (\sqrt{x} + \sqrt{7})$ $(sqrtx-sqrt7)(sqrtx+sqrt7)$ ?

2 Answers

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