frac{1}{sqrt{9}-sqrt{8}} = ? | vedclass

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

Question

(D) To solve the expression $\frac{1}{\sqrt{9}-\sqrt{8}}$,we rationalize the denominator.First,simplify the square roots: $\sqrt{9} = 3$ and $\sqrt{8}

Vedclass · Accepted Answer

$3+2\sqrt{2}$

Vedclass · Answer

(D) To solve the expression $\frac{1}{\sqrt{9}-\sqrt{8}}$,we rationalize the denominator.First,simplify the square roots: $\sqrt{9} = 3$ and $\sqrt{8} = \sqrt{4 	imes 2} = 2\sqrt{2}$.So,the expression becomes $\frac{1}{3-2\sqrt{2}}$.Now,multiply the numerator and the denominator by the conjugate $(3+2\sqrt{2})$:$\frac{1}{3-2\sqrt{2}} 	imes \frac{3+2\sqrt{2}}{3+2\sqrt{2}} = \frac{3+2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2}$.Calculate the denominator: $(3)^2 - (2\sqrt{2})^2 = 9 - (4 	imes 2) = 9 - 8 = 1$.Therefore,the final result is $\frac{3+2\sqrt{2}}{1} = 3+2\sqrt{2}$.

$\frac{1}{\sqrt{9}-\sqrt{8}} = ?$

Similar Questions

If $\sqrt{0.04 \times 0.4 \times a} = 0.4 \times 0.04 \times \sqrt{b},$ then the value of $\frac{a}{b}$ is

Given $\sqrt{2} = 1.414$. The value of $\sqrt{8} + 2\sqrt{32} - 3\sqrt{128} + 4\sqrt{50}$ is

The largest number of three digits which is a perfect square is

If $\sqrt{x} + \sqrt{441} = 0.02$,then the value of $x$ is

The least perfect square,which is divisible by $21, 36$ and $66$ completely is

Vedclass Test Series

Exam Paper Generator

Online Exam Module