Question

SIMPLIFY THE FOLLOWING BY RATIONALISING THE DENOMINATOR-
7 root 3 - 5 root 2/root 48 + root 18
PLEASE ANSWER IT.....WHOSEOVER WILL ANSWER FIRST I WILL MARK THE ANSWER BRAINLIEST...!

Answer 1

Question: Simplify the following by rationalizing the denominator.

$\implies \tt \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}}$

Solution:

$\Longrightarrow \ \tt \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18} \ }\\ \\ \\ \\\implies \tt \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18} \ } \times \dfrac{\left(\sqrt{48} - \sqrt{18}\right)}{\left(\sqrt{48} - \sqrt{18}\right)} \\ \\ \\ \\\Longrightarrow \ \tt \dfrac{\left(7\sqrt{3} - 5\sqrt{2}\right) \left(\sqrt{48} - \sqrt{18}\right)}{\left(\sqrt{48} + \sqrt{18}\right) \left(\sqrt{48} - \sqrt{18}\right)}$

Using the identity (a + b) (a - b) = a² - b² we get,

$\Longrightarrow \ \tt \dfrac{\left(7\sqrt{3} - 5\sqrt{2}\right) \left(\sqrt{48} - \sqrt{18}\right)}{\left(\sqrt{48}\right)^2 - \left(\sqrt{18}\right)^2}\\ \\ \\ \\\Longrightarrow \ \tt \dfrac{\left(7\sqrt{3} - 5\sqrt{2}\right) \left(\sqrt{48} - \sqrt{18}\right)}{48 - 18}$

$\left(\sf \sqrt{48} \ can \ be \ expressed \ as \ 4\sqrt{3}\right)\\ \\\left(\sf \sqrt{18} \ can \ be \ expressed \ as \ 3\sqrt{2} \right)$

$\Longrightarrow \ \tt \dfrac{\left(7\sqrt{3} - 5\sqrt{2}\right) \left(4\sqrt{3} - 3\sqrt{2}\right)}{30}\\ \\ \\ \\\Longrightarrow \ \tt \dfrac{(7\sqrt{3})(4\sqrt{3}) - (7\sqrt{3})(3\sqrt{2}) - (5\sqrt{2})(4\sqrt{3}) + (5\sqrt{2}) (3\sqrt{2})}{30}$

$\Longrightarrow \ \tt \dfrac{(28 \times 3) - (21 \times \sqrt{6}) - (20 \times \sqrt{6}) + (15 \times 2)}{30}\\ \\ \\ \\\Longrightarrow \ \tt \dfrac{84 \ - \ 21\sqrt{6} \ - \ 20\sqrt{6} \ + \ 30}{30}\\ \\ \\ \\\Longrightarrow \ \tt \dfrac{114 \ - \ 41\sqrt{6}}{30}$

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$\boxed{\sf Answer: \tt \dfrac{114 - 41\sqrt{6}}{30}}$