If frac{4 + 3sqrt{3}}{sqrt{7 + 4sqrt{3}}} = a | vedclass

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

Question

(C) Given the expression $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}} = a + \sqrt{b}$.First,simplify the denominator: $\sqrt{7 + 4\sqrt{3}} = \sqrt{7 +

Vedclass · Accepted Answer

$(-1, 12)$

Vedclass · Answer

(C) Given the expression $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}} = a + \sqrt{b}$.First,simplify the denominator: $\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + 2\sqrt{12}}$.Since $4 + 3 = 7$ and $4 	imes 3 = 12$,we can write this as $\sqrt{(\sqrt{4} + \sqrt{3})^2} = 2 + \sqrt{3}$.Now,the expression becomes $\frac{4 + 3\sqrt{3}}{2 + \sqrt{3}}$.Rationalize the denominator by multiplying the numerator and denominator by $(2 - \sqrt{3})$:$\frac{(4 + 3\sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{8 - 4\sqrt{3} + 6\sqrt{3} - 3(3)}{4 - 3} = \frac{8 + 2\sqrt{3} - 9}{1} = -1 + 2\sqrt{3}$.Rewrite $2\sqrt{3}$ as $\sqrt{2^2 	imes 3} = \sqrt{12}$.Thus,$-1 + \sqrt{12} = a + \sqrt{b}$.Comparing the terms,we get $a = -1$ and $b = 12$.Therefore,$(a, b) = (-1, 12)$.

If $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}} = a + \sqrt{b}$,then $(a, b) = $

Similar Questions

The rationalising factor of $2\sqrt{3} - \sqrt{7}$ is

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

The least number among $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}$ and $\sqrt[3]{8}$ is:

The value of $\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$ is

Evaluate: $\frac{4}{1 + \sqrt{2} - \sqrt{3}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module