Evaluate: frac{{[4 + sqrt{15}]}^{3/2} + {[4 - | vedclass

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

Question

(D) Let $x = 4 + \sqrt{15}$ and $y = 4 - \sqrt{15}$. Then $x+y = 8$ and $xy = 16 - 15 = 1$. We can write $x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqr

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $x = 4 + \sqrt{15}$ and $y = 4 - \sqrt{15}$. Then $x+y = 8$ and $xy = 16 - 15 = 1$. We can write $x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{2x} + \sqrt{2y})^3$. Since $2x = 8 + 2\sqrt{15} = (\sqrt{5} + \sqrt{3})^2$ and $2y = (\sqrt{5} - \sqrt{3})^2$,we have $\sqrt{2x} = \sqrt{5} + \sqrt{3}$ and $\sqrt{2y} = \sqrt{5} - \sqrt{3}$. Thus,$x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{5} + \sqrt{3} + \sqrt{5} - \sqrt{3})^3 = \frac{1}{\sqrt{2}} (2\sqrt{5})^3 = \frac{40\sqrt{5}}{\sqrt{2}} = 20\sqrt{10}$. Similarly,for the denominator,let $a = 6 + \sqrt{35}$ and $b = 6 - \sqrt{35}$. Then $2a = 12 + 2\sqrt{35} = (\sqrt{7} + \sqrt{5})^2$ and $2b = (\sqrt{7} - \sqrt{5})^2$. $a^{3/2} - b^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{2a}^3 - \sqrt{2b}^3) = \frac{1}{\sqrt{2}} ((\sqrt{7} + \sqrt{5})^3 - (\sqrt{7} - \sqrt{5})^3)$. Using $(u+v)^3 - (u-v)^3 = 2v^3 + 6u^2v$,we get $\frac{1}{\sqrt{2}} (2(5\sqrt{5}) + 6(7)\sqrt{5}) = \frac{1}{\sqrt{2}} (10\sqrt{5} + 42\sqrt{5}) = \frac{52\sqrt{5}}{\sqrt{2}} = 26\sqrt{10}$. The ratio is $\frac{20\sqrt{10}}{26\sqrt{10}} = \frac{20}{26} = \frac{10}{13}$.

Evaluate: $\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}$

Similar Questions

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

$0.4\overline{23} = ?$

The rational form of the number $1.\overline{41}$ is

If for $x \in \left(0, \frac{\pi}{2}\right)$,$\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$,$n > 0$,then the value of $n$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module