Question

IF X²+1/X²=18, THEN THE VALUE OF X+1/X WILL BE?

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Answer 1

Answer:

$\text{Given: } x^2 + \dfrac{1}{x^2} = 18\\\\\\\text{To Find: } x + \dfrac{1}{x} = ?\\\\\\\text{Squaring the "To Find" part we get:}\\\\\\\implies (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2\times x\times \dfrac{1}{x}\\\\\\\implies (x + \dfrac{1}{x})^2 = 18 + 2 ( 1 )\\\\\\\implies (x + \dfrac{1}{x})^2 = 20$

Taking square root on both sides we get:

$\implies (x + \dfrac{1}{x} ) = \sqrt{20} \\\\\implies \boxed{ \bf{(x + \dfrac{1}{x} ) = 2 \sqrt{5}}}$

Hence the required answer is 2√5.

Answer 2

Question:-

If $\tt \: { x }^{ 2 } + \dfrac { 1 } { { x }^{ 2 } } = 18$ , then the value of $\tt \: x + \dfrac { 1 } { x }$ will be ?

To Find:-

• Find the value of $\tt \: x + \dfrac { 1 } { x }$

Solution:-

Given ,

• $\tt \: { x }^{ 2 } + \dfrac { 1 } { { x }^{ 2 } } = 18$

$\tt\implies \: {( x + \dfrac { 1 } { x } )}^{ 2 } = { x }^{ 2 } + \dfrac { 1 } { { x }^{ 2 } } + 2 \times x \times \dfrac { 1 } {{ x }}$

$\tt\implies \: {( x + \dfrac { 1 } { x } )}^{ 2 } = 18 + 2( 1 )$

$\tt\implies \: x + \dfrac { 1 } { x } = \sqrt { 20 }$

$\implies {\pmb{\red{\sf \: x + \dfrac{ \: 1 \: }{ \: x \: } = 2 \sqrt{5} }}}$