Question

if x= 17+12 root 2, then the value of root x - 1/ root x is​

Answer 1

Answer: The value of √x - 1/√x is equal to 4√2

Step-by-step explanation:

$x = 17 + 12\sqrt{2}$

Expanding RHS,

⇒ $x = 9 + 8 +12\sqrt{2}$

⇒ $x = (3)^{2} + (2\sqrt{2} )^{2} +2(3)(2\sqrt{2})$

⇒ $x = (3 +2\sqrt{2} )^{2}$

Taking square root of both sides,

⇒ $\sqrt{x} = \sqrt{(3 + 2\sqrt{2} )^{2} }$

⇒ $\sqrt{x} = 3+2\sqrt{2}$               ...eq(i)

⇒ $\frac{1}{\sqrt{x} } = \frac{1}{3+2\sqrt{2} }$

⇒ $\frac{1}{\sqrt{x} } = \frac{1}{(3+2\sqrt{2}) } *\frac{(3-2\sqrt{2}) }{(3-2\sqrt{2} )}$

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

⇒ $\frac{1}{\sqrt{x} } = \frac{(3-2\sqrt{2}) }{(9-8 )}$

⇒ $\frac{1}{\sqrt{x} } = 3-2\sqrt{2}$      ...eq(ii)

Subtracting eq(ii) from eq(i),

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

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

$\sqrt{x} -\frac{1}{\sqrt{x} }= 4\sqrt{2}$  

#SPJ2