Question

If x = 9 + 4v5find the value of √x-1/√x​

Answer 1

Answer:

4

Step-by-step explanation:

→ x = 9 + 4√5

→ x = 5 + 4 + 4√5

→ x = (√5)² + (√4)² + 2*2*√5

→ x = (√5)² + (2)² + 2*2*√5

(a+b)² = a² + b² + 2*a*b

→ x = (√5 + 2)²

→ √x = √5 + 2

→ 1/√x = 1/(√5 + 2)

Multiply as well as divide by √5 - 2:

→ 1/√x = (√5 - 2)/(√5 + 2)(√5 - 2)

→ 1/√x = (√5 - 2)/(√5² - 2²)

→ 1/√x = (√5 - 2)/(5-4) = √5 - 2

Thus,

= > √x - 1/√x

= > √5 + 2 - (√5 - 2)

= > 4

Answer 2

ATQ:

x = 9 + 4√5

To find:

$\sf \Longrightarrow \sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }$

On Squaring we get:

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ } \Bigg\}^2 = \bigg\{\sqrt{x} \bigg\}^2 + \bigg\{\dfrac{1}{\sqrt{x}}\bigg\}^2 - 2\bigg\{\sqrt{x} \bigg\}\bigg\{\dfrac{1}{\sqrt{x}} \bigg\}$

Formula used:

(a - b)² = a² + b² - 2ab

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ } \Bigg\}^2 = \bigg\{\sqrt{9 + 4\sqrt{5 \ }}\bigg\}^2 + \Bigg\{\dfrac{1}{\sqrt{9 + 4\sqrt{5 \ }}}}\Bigg\}^2 - 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{1}{\Big\{\sqrt{9 + 4\sqrt{5}} \ \Big\}^2} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{1}{9 + 4\sqrt{5}} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{1}{9 + 4\sqrt{5}} \ \times \ \dfrac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} \ - \ 2$

Formula used:

(a + b) (a - b) = a² - b²

Where a = 9 and b = 4√5

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{9 - 4\sqrt{5}}{(9)^2 - (4\sqrt{5})^2} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{9 - 4\sqrt{5}}{81 - (16 \times 5)} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{9 - 4\sqrt{5}}{81 - 80} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ \dfrac{9 - 4\sqrt{5}}{1} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 + 4\sqrt{5} \ + \ 9 - 4\sqrt{5} \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 9 \ + \ 9 \ - \ 2$

$\sf \Longrightarrow \Bigg\{\sqrt{x} \ - \ \dfrac{1}{\sqrt{x} \ }\Bigg\}^2 = 16$

$\sf \Longrightarrow \sqrt{x} \ - \dfrac{1}{\sqrt{x}} =\sqrt{16}$

$\sf \Longrightarrow \sqrt{x} \ - \dfrac{1}{\sqrt{x}} = 4$

Therefore √x - (1/√x) is 4.