How do you simplify # (2+sqrt(x)) / (sqrt(2 x)+sqrt(8)) #?

2 Answers

Answer:

#sqrt2#

Explanation:

When working a problem such as this, keep in mind that if we take something (like our denominator) that has the form of #a+b#, we can multiply by #a-b# and we'll end up with #a^2-b^2#. And so:

#(2+sqrtx)/(sqrt(2x)+sqrt8)(1)#

#(2+sqrtx)/(sqrt(2x)+sqrt8)((sqrt(2x)-sqrt8)/(sqrt(2x)-sqrt8))#

#((2+sqrtx)(sqrt(2x)-sqrt8))/((sqrt(2x)+sqrt8)(sqrt(2x)-sqrt8))#

#(2sqrt(2x)-2sqrt8+sqrtxsqrt(2x)-sqrtxsqrt8)/(2x-8)#

Now for some simplification in the numerator. Remember that #sqrt8=2sqrt2#:

#(2sqrt(2x)-2(2sqrt2)+sqrt(2x^2)-2sqrt(2x))/(2x-8)#

#(-4sqrt2+xsqrt2)/(2x-8)#

#(sqrt2(x-4))/(2(x-4))#

#sqrt2/2#

Remember that #(pmsqrt2)^2=2#

#sqrt2/(sqrt2)^2=sqrt2#

and

#sqrt2/(-sqrt2)^2=sqrt2#

Jul 2, 2018

Answer:

#sqrt(2)/2 = 1/sqrt(2)#

Explanation:

Given: #(2 + sqrt(x))/(sqrt(2x) + sqrt(8))#

Multiply both the numerator and the denominator by the conjugate of the denominator, which is basically multiplying the expression by #1#.

The conjugate is a factor that when multiplied, cancels the middle term. The conjugate of #sqrt(2x) + sqrt(8)# is #sqrt(2x) - sqrt(8).#

#(2 + sqrt(x))/(sqrt(2x) + sqrt(8)) * (sqrt(2x) - sqrt(8))/(sqrt(2x) - sqrt(8)) #

#= (2 sqrt(2x)-2 sqrt(8) + sqrt(x) sqrt(2x) - sqrt(x)sqrt(8))/(sqrt(2x) sqrt(2x) - sqrt(8)sqrt(8))#

Use the Radical rules: #sqrt(a)sqrt(b) = sqrt(ab); " "sqrt(x)sqrt(x) = sqrt(x^2) = x#

Simplify #sqrt(8): " "sqrt(8) = sqrt(4*2) = sqrt(4)sqrt(2) = 2 sqrt(2)#

#=(cancel(2 sqrt(2x))-2*2 sqrt(2) + x sqrt(2) cancel(- 2 sqrt(2x)))/(sqrt(4x^2) - sqrt(64)) #

#= (-4sqrt(2) + xsqrt(2))/(2x - 8) = (xsqrt(2) - 4sqrt(2))/(2x - 8)#

Factor: #" " (sqrt(2)cancel(x - 4))/(2cancel(x - 4))#

#(2 + sqrt(x))/(sqrt(2x) + sqrt(8))= sqrt(2)/2#