How do you simplify #(sqrtx-sqrt7)(sqrtx+sqrt7)#?

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Answer 1 · 2016-09-05T17:17:08

#x - 7#

We have: #(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))#

#= x + sqrt(7 x) - sqrt(7 x) - 7#

#= x - 7#

Answer 2 · 2016-09-05T18:39:21

#x-7#

Recall: difference of squares can be factored:

#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#

This is what we have in this question,

#(sqrtx -sqrt7)(sqrtx + sqrt7) = (sqrtx)^2 - (sqrt7)^2#

=#x-7#