Question

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

Answer 1

x=7+√40

x=7+2√10
  =(√5)²+(√2)²+2×√2×√5
  =(√5+√2)²

√x=√5+√2
1/√x=1/√5+√2
      =√5-√2/3

√x+1/√x=√5+√2 + √5-√2/3
            =4√5+2√2/3
            ie 2/3( 2√5+√2)

hope this is helpful

Answer 2

Answer:

Required value of

$\sqrt{x} + \frac{1}{ \sqrt{x} }$ is $\frac{2}{3}(2\sqrt{5}+\sqrt{2})$

Step-by-step explanation:

Given,

$x = 7 + \sqrt{40} \\ x = 7 + \sqrt{2 \times 2 \times 2 \times 5} \\ x = 7 + 2 \sqrt{2 \times 5} \\ x = 7 + 2 \sqrt{10}$

Now we can write

$7 + 2 \sqrt{10} \\ = 5 + 2 + 2 \sqrt{10} \\ = { \sqrt{5} }^{2} + { \sqrt{2} }^{2} + 2 \times \sqrt{5} \times \sqrt{2} \\ = {( \sqrt{5} + \sqrt{2} ) }^{2}$

So,

$x = {( \sqrt{5} + \sqrt{2} ) }^{2} \\ \sqrt{x} = \sqrt{5} + \sqrt{2}$

and

$\frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + \sqrt{2} } \\ = \frac{ \sqrt{5} - \sqrt{2} }{( \sqrt{5} + \sqrt{2} )( \sqrt{5} - \sqrt{2}) } \\ = \frac{ \sqrt{5} - \sqrt{2} }{ { \sqrt{5} }^{2} - { \sqrt{2} }^{2} } \\ = \frac{ \sqrt{5} - \sqrt{2} }{5 - 2} \\ = \frac{ \sqrt{5} - \sqrt{2} }{3}$

So,

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

Here applied formulas are

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ {a}^{2} - {b}^{2} = (a + b)(a - b)$

This is a problem of Power of indices.

Some important formulas of Power of indices:

${x}^{0} = 1 \\ {x}^{1} = x \\ {x}^{a} \times {x}^{b} = {x}^{a + b} \\ \frac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b} \\ {x}^{ {y}^{a} } = {x}^{ya} \\ {x}^{ - 1} = \frac{1}{x} \\ {x}^{a} \times {y}^{a} = {(xy)}^{a}$

Power of indices related two more questions:

https://brainly.in/question/20611233

https://brainly.in/question/8929724

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