Question

if x=9 +4root 5then find the value of x +1/x.

plz answer now​

Answer 1

Given

$\tt \to \: x = 9 + 4 \sqrt{5}$

To Find Value of

$\tt \to \: x + \dfrac{1}{x}$

If

$\tt \to \: x = 9 + 4 \sqrt{5}$

Then

$\tt \to \: \dfrac{1}{x} = \dfrac{1}{9 + 4 \sqrt{5} }$

Now Rationalize the Denominator

$\tt \to \: \dfrac{1}{x} = \dfrac{1}{9 + 4 \sqrt{5} } \times \dfrac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} }$

Using this identity

$\tt \to(a + b)(a - b) = {a}^{2} - {b}^{2}$

we get

$\tt \to \: \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{(9) {}^{2} - (4 \sqrt{5}) }$

$\tt \to \: \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{81 - 16 \times 5}$

$\tt \to \: \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{81 - 80} = 9 - 4 \sqrt{5}$

Now we have to find

$\tt \to \: x + \dfrac{1}{x}$

Put the value

$\tt \to \: 9 + 4 \sqrt{5} + 9 - 4 \sqrt{5}$

$\tt \to \: 9 + 9 = 18$

Answer

$\tt \to \: x + \dfrac{1}{x} = 18$

Answer 2

$\dag{\underline{\sf{\pmb{Given: - }}}}$

$\\\sf{\rightarrow{x = 9 + 4 \sqrt{5}}}$

$\\\dag{\underline{\sf{\pmb{To \: Find: - }}}}$

$\\\sf{\rightarrow{Value \: of \: \bold{x}}}$

$\\\dag{\underline{\sf{\pmb{Solution}}}}$

$\\ \sf{\dfrac{1}{x} = \dfrac{1}{9 + 4 \sqrt{5}}}$

$\\ \sf{\implies{\dfrac{1}{x} = \dfrac{1}{9 + 4 \sqrt{5} } \times \dfrac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\sf{Rationalizing \: the \: Denominator}}$

$\\ \sf{\implies{ \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{(9) {}^{2} - {(4 \sqrt{5}) }^{2}}}} \: \: \: \: \: \: \: \: \boxed{\sf{\because{(a + b)(a - b) = {a}^{2} - {b}^{2}}}}$

$\\ \sf{\implies{ \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{81 - 16 \times 5}}}$

$\\ \sf{\implies{ \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{81 - 80}}}$

$\\\\ \sf{\implies{ \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5} }{1}}}$

$\\\sf{\implies{ \dfrac{1}{x} = \bold{9 - 4\sqrt{5}}}}$

Now, our given equation is :-

$\\ \sf{\bold{x + \dfrac{1}{x}}}$

Putting the values:-

$\\ \sf{9 + 4 \sqrt{5} + 9 - 4 \sqrt{5}}$

$\\ \sf{\implies{9 + \cancel{4 \sqrt{5}} + 9 - \cancel{4 \sqrt{5}}}}$

$\\ \sf{\implies{\bold{\red{18}}}}$

$\\ \underline{\boxed{\rm{Hence, the \: value \: of \: \bold{x + \dfrac{1}{x}} \: is \; \bold{18}}}}$