Question

If x=7+√40 , find the value of √x + (1/√x)

Answer 1

Answer in attachment. Answer is 4√5+2√2/3. Please comment if you need detailed explanation. Try to solve such problems by yourself in the future. :)

Answer 2

Answer:

$\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{4\sqrt{5}+2\sqrt{2}}{3}$

Step-by-step explanation:

i) Given x = 7+√40

=> x = 7+√2×2×10

=> x = 7+2√10

=> x = 7+2×√5×√2

=> x = 5+2+2×√5×√2

=> x = (√5)²+(√2)²+2×√5×√2

=> x = (√5+√2)²

=> √x = √5+√2 ----(1)

Now ,

$ii)\frac{1}{\sqrt{x}}$

= $\frac{1}{\sqrt{5}+\sqrt{2}}$

/* Rationalising the denominator, we get

=$\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$

=$\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}$

=$\frac{\sqrt{5}-\sqrt{2}}{5-2}$

=$\frac{\sqrt{5}-\sqrt{2}}{3}$ ----(2)

$iii) \sqrt{x}+\frac{1}{\sqrt{x}}$

=$\sqrt{5}+\sqrt{2}+\frac{\sqrt{5}-\sqrt{2}}{3}$

=$\frac{3(\sqrt{5}+\sqrt{2})+\sqrt{5}-\sqrt{2}}{3}$

=$\frac{3\sqrt{5}+3\sqrt{2}+\sqrt{5}-\sqrt{2}}{3}$

=$\frac{4\sqrt{5}+2\sqrt{2}}{3}$

Therefore,

$\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{4\sqrt{5}+2\sqrt{2}}{3}$

•••♪