Question

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

Answer 1

`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`

Answer 2

`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`