JIPMAT 2022 (QA) - sqrt(5)+sqrt(3)8-2 sqrt(15)+11+2 sqrt(30)sqrt(6)-sqrt(5) | PYQs + Solutions

JIPMAT 2022

Algebra

Indices

Hard

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}+\frac{\sqrt{11+2 \sqrt{30}}}{\sqrt{6}-\sqrt{5}}$

25+\sqrt{10}+2 \sqrt{30}

40+\sqrt{10}+2 \sqrt{30}

30+\sqrt{20}+2 \sqrt{30}

15+\sqrt{15}+2 \sqrt{30}

Correct Option: 4

Look at the first term first:

\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}

If you look closely, you'll see that:

8-2\sqrt{15}=(\sqrt{5})^2+(\sqrt{3})^2-2(\sqrt5)(\sqrt3)

We know that

(a-b)^2=a^2+b^2-2ab

So we can write the above expression as

(\sqrt5-\sqrt3)^2

So, the first term:

\frac{\sqrt{5}+\sqrt{3}}{\sqrt5-\sqrt3}

Multiply numerator and denominator by

(\sqrt{5}+\sqrt3)

\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)}

We know that

(a+b)(a-b)=a^2-b^2

and that

(a+b)^2=a^2+b^2+2ab

So the above expression can be written as:

\frac{5+3+2\sqrt15}{5-3}=\frac{5+3+2\sqrt{15}}{2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}

Now for the second term, it's the same approach.

11+2\sqrt{30}=(\sqrt6)^2+(\sqrt5)^2+2(\sqrt6)(\sqrt5)=(\sqrt6+\sqrt5)^2

So, second term:

\frac{\sqrt6+\sqrt5}{\sqrt{6}-\sqrt{5}}

Multiply numerator and denominator by

(\sqrt6+\sqrt5)

\frac{(\sqrt6+\sqrt5)^2}{(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}

Open the brackets using the same identity

\frac{6+5+2\sqrt{30}}{6-5}=\frac{6+5+2\sqrt{30}}{1}=11+2\sqrt{30}

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Just add the two terms now, and we'll get

4+\sqrt{15}+11+2\sqrt{30}=15+\sqrt{15}+2\sqrt{30}