3 years ago

Wanna play a game on vc

Mathematics

1 answer:

andrey2020 [161]3 years ago

7 0

Yes why not what’s your username

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Read 2 more answers

What is the following quotient?

seropon [69]

Answer: Choice B) $\frac{5\sqrt{11}+5\sqrt{3}}{8}\\\\$

==========================================

Work Shown:

$\frac{5}{\sqrt{11}-\sqrt{3}}\\\\\\\frac{5(\sqrt{11}+\sqrt{3})}{(\sqrt{11}-\sqrt{3})(\sqrt{11}+\sqrt{3})}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{(\sqrt{11})^2-(\sqrt{3})^2}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{11-3}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{8}\\\\\\$

This shows why choice B is the answer.

---------------

Explanation:

In the second step, I multiplied top and bottom by $(\sqrt{11}+\sqrt{3})$
This is so we can apply the difference of squares rule in step 3. The difference of squares rule is (x-y)(x+y) = x^2-y^2.
In step 4, the square roots cancel out with the squaring operation. The two operations are inverses of one another, which is why they cancel.

So in general, if the denominator is $\sqrt{a}-\sqrt{b}$ and you want to rationalize the denominator, then you should multiply top and bottom by $\sqrt{a}+\sqrt{b}$ . The same applies in reverse as well.

This leads to the denominator becoming $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = (\sqrt{a})^2-(\sqrt{b})^2 = a-b$

Keep in mind that $a \ge 0$ and $b \ge 0$

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