Question

if x= 3- root 3 then find the value of root x -1/ root x​

Answer 1

Answer:

hen find the value of Given, ⇒ x = 2 + 3 + 2√6 ⇒ x = (√2) 2 +

If then find the value of Given, ⇒ x = 2 + 3 + 2√6 ⇒ x = (√2) 2 +

| If then find the value of

A. 4

B. √3

C. 2√3

D. 0

Answer 2

given

$x = 3 - \sqrt{3}$

so we can write

$\frac{1}{x} = \frac{1}{3 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{1}{3 - \sqrt{3} } \times \frac{3 + \sqrt{3} }{3 + \sqrt{3} } \\ \\ \frac{1}{x} = \frac{3 + \sqrt{3} }{9 - 3} = \frac{3 + \sqrt{3} }{6} \\ \\ \frac{1}{x} = \frac{1}{2} + \frac{ \sqrt{3} }{6}$

Now

$x + \frac{1}{x} = 3 - \sqrt{3} + \frac{1}{2} + \frac{ \sqrt{3} }{6} \\ \\ x + \frac{1}{x} = \frac{7}{2} - \frac{5 \sqrt{3} }{6} \\ \\ x + \frac{1}{x} - 2 = \frac{7}{2} - \frac{5 \sqrt{3} }{6} - 2 \\ \\ ( { \sqrt{x} - \frac{1}{ \sqrt{x} } })^{2} = \frac{3}{2} - \frac{ 5\sqrt{3} }{6}$

$\sqrt{x} - \frac{1}{ \sqrt{x} } = \sqrt{ \frac{3 }{2} - \frac{5 \sqrt{3} }{6} }$