How do you simplify $\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}}$ $(2+sqrt(x)) / (sqrt(2 x)+sqrt(8))$ ?

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How do you simplify $\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}}$ $(2+sqrt(x)) / (sqrt(2 x)+sqrt(8))$ ?

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Answer 1 · 2018-07-02T03:47:47

$\sqrt{2}$ $sqrt2$

Explanation:

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

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} (1)$ $(2+sqrtx)/(sqrt(2x)+sqrt8)(1)$

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} (\frac{\sqrt{2 x} - \sqrt{8}}{\sqrt{2 x} - \sqrt{8}})$ $(2+sqrtx)/(sqrt(2x)+sqrt8)((sqrt(2x)-sqrt8)/(sqrt(2x)-sqrt8))$

$\frac{(2 + \sqrt{x}) (\sqrt{2 x} - \sqrt{8})}{(\sqrt{2 x} + \sqrt{8}) (\sqrt{2 x} - \sqrt{8})}$ $((2+sqrtx)(sqrt(2x)-sqrt8))/((sqrt(2x)+sqrt8)(sqrt(2x)-sqrt8))$

$\frac{2 \sqrt{2 x} - 2 \sqrt{8} + \sqrt{x} \sqrt{2 x} - \sqrt{x} \sqrt{8}}{2 x - 8}$ $(2sqrt(2x)-2sqrt8+sqrtxsqrt(2x)-sqrtxsqrt8)/(2x-8)$

Now for some simplification in the numerator. Remember that $\sqrt{8} = 2 \sqrt{2}$ $sqrt8=2sqrt2$ :

$\frac{2 \sqrt{2 x} - 2 (2 \sqrt{2}) + \sqrt{2 x^{2}} - 2 \sqrt{2 x}}{2 x - 8}$ $(2sqrt(2x)-2(2sqrt2)+sqrt(2x^2)-2sqrt(2x))/(2x-8)$

$\frac{- 4 \sqrt{2} + x \sqrt{2}}{2 x - 8}$ $(-4sqrt2+xsqrt2)/(2x-8)$

$\frac{\sqrt{2} (x - 4)}{2 (x - 4)}$ $(sqrt2(x-4))/(2(x-4))$

$\frac{\sqrt{2}}{2}$ $sqrt2/2$

Remember that ${(\pm \sqrt{2})}^{2} = 2$ $(pmsqrt2)^2=2$

$\frac{\sqrt{2}}{{(\sqrt{2})}^{2}} = \sqrt{2}$ $sqrt2/(sqrt2)^2=sqrt2$

and

$\frac{\sqrt{2}}{{(- \sqrt{2})}^{2}} = \sqrt{2}$ $sqrt2/(-sqrt2)^2=sqrt2$

Answer 2 · 2018-07-02T03:53:18

$\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ $sqrt(2)/2 = 1/sqrt(2)$

Explanation:

Given: $\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}}$ $(2 + sqrt(x))/(sqrt(2x) + sqrt(8))$

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

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

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} \cdot \frac{\sqrt{2 x} - \sqrt{8}}{\sqrt{2 x} - \sqrt{8}}$ $(2 + sqrt(x))/(sqrt(2x) + sqrt(8)) * (sqrt(2x) - sqrt(8))/(sqrt(2x) - sqrt(8))$

$= \frac{2 \sqrt{2 x} - 2 \sqrt{8} + \sqrt{x} \sqrt{2 x} - \sqrt{x} \sqrt{8}}{\sqrt{2 x} \sqrt{2 x} - \sqrt{8} \sqrt{8}}$ $= (2 sqrt(2x)-2 sqrt(8) + sqrt(x) sqrt(2x) - sqrt(x)sqrt(8))/(sqrt(2x) sqrt(2x) - sqrt(8)sqrt(8))$

Use the Radical rules: $\sqrt{a} \sqrt{b} = \sqrt{a b}; \sqrt{x} \sqrt{x} = \sqrt{x^{2}} = x$ $sqrt(a)sqrt(b) = sqrt(ab); " "sqrt(x)sqrt(x) = sqrt(x^2) = x$

Simplify $\sqrt{8} : \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \sqrt{2} = 2 \sqrt{2}$ $sqrt(8): " "sqrt(8) = sqrt(4*2) = sqrt(4)sqrt(2) = 2 sqrt(2)$

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

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

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

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

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