How do you rationalize the denominator and simplify $\frac{4}{\sqrt{7} - \sqrt{5}}$ $4/(sqrt7-sqrt5)$ ?

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How do you rationalize the denominator and simplify $\frac{4}{\sqrt{7} - \sqrt{5}}$ $4/(sqrt7-sqrt5)$ ?

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Answer 1 · 2017-11-12T18:01:08

$= 2 (\sqrt{7} + \sqrt{5})$ $=2(sqrt7+sqrt5)$

Explanation:

$\frac{4}{\sqrt{7} - \sqrt{5}}$ $4/(sqrt7-sqrt5)$

to rationalise the denominator we make use of

$(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b)=a^2-b^2$

so

$\frac{4}{\sqrt{7} - \sqrt{5}} = \frac{4}{\sqrt{7} - \sqrt{5}} \times \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}$ $4/(sqrt7-sqrt5)= 4/(sqrt7-sqrt5)xx (sqrt7+sqrt5)/(sqrt7+sqrt5)$

$= \frac{4 (\sqrt{7} + \sqrt{5})}{{(\sqrt{7})}^{2} - {(\sqrt{5})}^{2}}$ $=(4(sqrt7+sqrt5))/((sqrt7)^2-(sqrt5)^2)$

$= \frac{4 (\sqrt{7} + \sqrt{5})}{7 - 5}$ $=(4(sqrt7+sqrt5))/(7-5)$

$=(cancel(4)^2(sqrt7+sqrt5))/cancel(2)$

$=2(sqrt7+sqrt5)$

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How do you rationalize the denominator and simplify $\frac{4}{\sqrt{7} - \sqrt{5}}$ $4/(sqrt7-sqrt5)$ ?

1 Answer

Explanation:

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How do you rationalize the denominator and simplify $\frac{4}{\sqrt{7} - \sqrt{5}}$ $4/(sqrt7-sqrt5)$ ?