Question

rationalize the denominator of 3 root 5 + root 3/root 5 - root 3

Answer 1

Answer:

By rationalizing the denominator of $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ get the answer as $\bold{9+2 \sqrt{15}}$

Step-by-step explanation:

Rationalizing the denominator means multiplying the numerator in a fraction by the value of the denominator but by its opposite signed value.

Now to rationalize the above figure we multiply both the numerator and denominator by [√5+√3]

$\left[\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\right]\left[\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right]$

On removing the brackets and multiplying,

$=\frac{3 \sqrt{5}+\sqrt{3} \cdot \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3} \cdot \sqrt{5}+\sqrt{3}}$

The denominator is multiplied by using the formula $\bold{(a+b)(a-b)=a^{2}-b^{2}}$

Taking a=√5 and b=√3

$\frac{3 \sqrt{5}+\sqrt{3} \cdot \sqrt{5}+\sqrt{3}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}=\frac{3 \sqrt{5}+\sqrt{3} \cdot \sqrt{5}+\sqrt{3}}{5-3}$

$\begin{array}{l}{\frac{3 \sqrt{5}+\sqrt{3} \cdot \sqrt{5}+\sqrt{3}}{2}=\frac{18+4 \sqrt{15}}{2}} \\ {\frac{18}{2}+\frac{4 \sqrt{15}}{2}=9+2 \sqrt{15}}\end{array}$

By rationalizing the denominator of $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ get the answer as $\bold{9+2 \sqrt{15}}$

Answer 2

Step-by-step explanation:

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