If x = 3 - sqrt{5},then find the value of fra | vedclass

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

Question

(B) Given $x = 3 - \sqrt{5}$.First,calculate $3x - 2$:$3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.To simplify $\sqrt{7 - 3\sqrt

Vedclass · Accepted Answer

$1/\sqrt{5}$

Vedclass · Answer

(B) Given $x = 3 - \sqrt{5}$.First,calculate $3x - 2$:$3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.To simplify $\sqrt{7 - 3\sqrt{5}}$,multiply and divide by $\sqrt{2}$:$\sqrt{7 - 3\sqrt{5}} = \sqrt{\frac{14 - 6\sqrt{5}}{2}} = \frac{\sqrt{14 - 2\sqrt{45}}}{\sqrt{2}}$.Since $9 + 5 = 14$ and $9 	imes 5 = 45$,we have $\sqrt{14 - 2\sqrt{45}} = \sqrt{9} - \sqrt{5} = 3 - \sqrt{5}$.So,$\sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}}$.Now substitute this into the expression:$\frac{\sqrt{x}}{\sqrt{2} + \frac{3 - \sqrt{5}}{\sqrt{2}}} = \frac{\sqrt{x}}{\frac{2 + 3 - \sqrt{5}}{\sqrt{2}}} = \frac{\sqrt{2} \cdot \sqrt{x}}{5 - \sqrt{5}}$.Since $x = 3 - \sqrt{5}$,$\sqrt{2} \cdot \sqrt{x} = \sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5} - 1)^2} = \sqrt{5} - 1$.Substituting back: $\frac{\sqrt{5} - 1}{5 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

If $x = 3 - \sqrt{5}$,then find the value of $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}}$.

Similar Questions

If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$,then the value of $x$ is equal to

If ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$,then $x =$

$\frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}} = $

If $x = 2 + \sqrt{3}$ and $xy = 1$,then find the value of $\frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}}$.

The solution to the equation $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module