Question

If x=7+root 40 find, rootx +root 1/x

Answer 1

x=7+√10
x = 7+√4×10
x = 7+2√10
then,
√x + (√1/x) = √(7+2√10) + (√1/(7+2√10))
   = √(7+2√10) + (1/(7+2√10))            LCM
  =  [√(7+2√10)][√(7+2√10)] +1 / √(7+2√10)
   = 7+2√10+1/ √(7+2√10)
  =  8+2√10 /√(7+2√10)

Answer 2

Answer:

$\sqrt{x}+\sqrt{\frac{1}{x}}=\frac{8+2\sqrt{10}}{\sqrt{7+2\sqrt{10}}}$

Step-by-step explanation:

Given : $x= 7+\sqrt{40}$

To Find : $\sqrt{x}+\sqrt{\frac{1}{x}}$

Solution :

$x= 7+\sqrt{40}$

$x= 7+\sqrt{2 \times 2 \times 10}$

$x= 7+2\sqrt{10}$

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

$=\sqrt{ 7+2\sqrt{10}}+\sqrt{\frac{1}{ 7+2\sqrt{10}}}$

$=\frac{7+2\sqrt{10}+1}{\sqrt{7+2\sqrt{10}}}$

$=\frac{8+2\sqrt{10}}{\sqrt{7+2\sqrt{10}}}$

Hence $\sqrt{x}+\sqrt{\frac{1}{x}}=\frac{8+2\sqrt{10}}{\sqrt{7+2\sqrt{10}}}$