If x = frac{sqrt{5} + sqrt{2}}{sqrt{5} - sqrt | vedclass

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

Question

(B) Given $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.First,calculate $x + y$ and

Vedclass · Accepted Answer

$\frac{1}{3}[56\sqrt{10} + 12]$

Vedclass · Answer

(B) Given $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.First,calculate $x + y$ and $xy$:$x + y = \frac{(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(5 + 2 + 2\sqrt{10}) + (5 + 2 - 2\sqrt{10})}{5 - 2} = \frac{14}{3}$.$xy = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} 	imes \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} = 1$.Now,$3x^2 - 3y^2 + 4xy = 3(x - y)(x + y) + 4xy$.We know $(x - y)^2 = (x + y)^2 - 4xy = (\frac{14}{3})^2 - 4(1) = \frac{196}{9} - 4 = \frac{196 - 36}{9} = \frac{160}{9}$.So,$x - y = \frac{\sqrt{160}}{3} = \frac{4\sqrt{10}}{3}$.Substituting these values into the expression:$3(\frac{4\sqrt{10}}{3})(\frac{14}{3}) + 4(1) = \frac{56\sqrt{10}}{3} + 4 = \frac{56\sqrt{10} + 12}{3} = \frac{1}{3}[56\sqrt{10} + 12]$.

If $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$,then find the value of $3x^2 + 4xy - 3y^2$.

Similar Questions

Find the solution of the equation $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$.

The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2 + 5x + 3) = 4 - \log_{(2x+3)}(x^2 + 2x + 1)$ is equal to . . . . . . .

If ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,and ${a^z} = {(x + y + z)^x}$,then:

Find the value of $x$ in the equation $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$.

The value of ${81^{(1/{\log_5}3)}} + {27^{\log_9 36}} + {3^{4/{\log_7}9}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module