Simplify: frac{3+sqrt{6}}{17 sqrt{3}-2 sqrt{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given expression: $\frac{3+\sqrt{6}}{17 \sqrt{3}-2 \sqrt{32}+3 \sqrt{18}-4 \sqrt{48}}$Simplify the denominator terms:$2 \sqrt{32} = 2 \sqrt{16 	i

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) Given expression: $\frac{3+\sqrt{6}}{17 \sqrt{3}-2 \sqrt{32}+3 \sqrt{18}-4 \sqrt{48}}$Simplify the denominator terms:$2 \sqrt{32} = 2 \sqrt{16 	imes 2} = 2 	imes 4 \sqrt{2} = 8 \sqrt{2}$$3 \sqrt{18} = 3 \sqrt{9 	imes 2} = 3 	imes 3 \sqrt{2} = 9 \sqrt{2}$$4 \sqrt{48} = 4 \sqrt{16 	imes 3} = 4 	imes 4 \sqrt{3} = 16 \sqrt{3}$Substitute these back into the denominator:$17 \sqrt{3} - 8 \sqrt{2} + 9 \sqrt{2} - 16 \sqrt{3}$Group like terms:$(17 \sqrt{3} - 16 \sqrt{3}) + (9 \sqrt{2} - 8 \sqrt{2}) = \sqrt{3} + \sqrt{2}$Now the expression becomes: $\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}$Factor the numerator: $3 + \sqrt{6} = \sqrt{3} 	imes \sqrt{3} + \sqrt{3} 	imes \sqrt{2} = \sqrt{3}(\sqrt{3} + \sqrt{2})$Substitute back: $\frac{\sqrt{3}(\sqrt{3} + \sqrt{2})}{\sqrt{3} + \sqrt{2}} = \sqrt{3}$

Simplify: $\frac{3+\sqrt{6}}{17 \sqrt{3}-2 \sqrt{32}+3 \sqrt{18}-4 \sqrt{48}}$

Similar Questions

Express $7429$ as a product of primes.

According to Euclid's division lemma,for given positive integers $a$ and $7$,there exist unique non-negative integers $q$ and $r$ such that $a = 7q + r$; where..........

The values of the remainder $r$,when a positive integer $a$ is divided by $3$ are $0$ and $1$ only. Justify your answer.

Write whether every positive integer can be of the form $4q + 2$,where $q$ is an integer. Justify your answer.

Find the square root of: $\frac{7}{4} + \sqrt{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module