Question

Hii Brainlians


queston :- If x = 5 +2√6 then find the value of
√x+ 1/√x and √x - 1/√x


Please Give correct answer​

Answer 1

Step-by-step explanation:

Given: x = 5 + 2√6

⇒ 3 + 2 + 2√6

⇒ (√3)² + (√2)² + 2(√3)(√2)

⇒ (√3 + √2)²

So, x = (√3 + √2)²

Then, √x = (√3 + √2)

Now,

(1/√x) = (1/√3 + √2) * [(√3 - √2)/(√3 - √2)]

          = (√3 - √2)/(√3)² - (√2)²

          = (√3 - √2)

(i)

√x + (1/√x) = √3 + √2 + √3 - √2

                  = 2√3

(ii)

√x - (1/√x) = √3 + √2 - √3 + √2

                 = 2√2

Hope it helps!

Answer 2

SOLUTION ☺️

Given: x= 5+2√6

1st case= √x+1/√x

$= > \frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } \\ \\ = > \frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } = \frac{5 - 2 \sqrt{6} }{25 - 24} = 5 - 2 \sqrt{6} \\ \\ = > now \\ ( \sqrt{x} \times \frac{1}{ \sqrt{x} } ) {}^{2} = x + \frac{1}{x} + 2 \times \sqrt{x} \times \frac{1}{ \sqrt{x} } = x + \frac{1}{x} + 2 \\ \\ = > \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{x + \frac{1}{x} + 2 } \\ \\ = > \sqrt{5 + 2} \sqrt{6 + 5 - 2 \sqrt{6 + 2} } \\ \\ = > \sqrt{5 + 5 + 2} \\ \\ = > \sqrt{12}$

2nd Case

= √x-1/√x

$given > x = 5 + 2 \sqrt{6} \\ \\ = > \frac{1}{x} = 5 - 2 \sqrt{6} \: \: rationalise \\ \\ = > now \: \\ ( \sqrt{x} - \frac{1}{ \sqrt{x} } ) {}^{2} = x + \frac{1}{x} - 2 \times \sqrt{x} \times \frac{1}{x} = x + \frac{1}{x} - 2 \\ \\ = > \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{x + \frac{1}{x} - 2 } \\ \\ = > \sqrt{5 + 2 \sqrt{6 + 5 - 2 \sqrt{6 - 2} } } \\ \\ = > \sqrt{5 + 5 - 2} \\ \\ = > \sqrt{10}$

HOPE IT HELPS