Question

x=17+12root2 then rootx-1/root x is

Answer 1

Answer:

$4 \sqrt{2}$

Step-by-step explanation:

Given question:

$x = 17 + 12 \sqrt{2} \\ find \sqrt{x} - \frac{1}{ \sqrt{x} }$

Solution:

$let \: y = \sqrt{x} - \frac{1}{ \sqrt{x} } \\ = > {y}^{2} = ( \sqrt{x} - \frac{1}{ \sqrt{x} } )^{2} \\ = > {y}^{2} = ( \sqrt{x})^{2} + {( \frac{1}{ \sqrt{x} } })^{2} - (2 \times \sqrt{x} \times \frac{1}{ \sqrt{x} } ) \\ = > {y}^{2} = x + \frac{1}{x} - 2 \\ = > {y }^{2} = \frac{ {x}^{2} - 2x + 1}{x} \\ = > y {}^{2} = \frac{ {(x - 1)}^{2} }{x} \\ = > y = \sqrt{\frac{ {(x - 1)}^{2} }{x} } \\ = > y = \frac{x - 1}{ \sqrt{x} }$

We can write the value of X as:

$x = 17 + 12 \sqrt{2} \\ = 9 + 8 + 12 \sqrt{2} \\ = 9 + 8 + 2 \times 3 \times 2 \sqrt{2} \\ = {3}^{2} + (2 \sqrt{2} ) {}^{2} + (2 \times 3 \times 2 \sqrt{2} ) \\ = (3 + 2 \sqrt{2} ) ^{2} \\ \\ so \sqrt{x} = 3 + 2 \sqrt{2}$

So, by placing the value,

$y = \frac{x - 1}{ \sqrt{x} } \\ = > y = \frac{17 + 12 \sqrt{2} - 1 }{3 + 2 \sqrt{2} } \\ = > y = \frac{16 + 12 \sqrt{2} }{3 + 2 \sqrt{2} } \\ rationalizing \: the \: denomintor \\ = > y = \frac{(16 + 12 \sqrt{2})(3 - 2 \sqrt{2} ) }{3 {}^{2} - (2 \sqrt{2}) {}^{2} } \\ = > y = \frac{48 - 32 \sqrt{2 } + 36 \sqrt{2} - 48}{9 - 8} \\ = > y = 4 \sqrt{2} \: \: (ans.)$

Hope this helps. ^_^