Question

if X square + 1 upon x square equal to 51 then find value of x minus one upon x

Answer 1

$(x - \frac{1}{x} ) {}^{2} \\ = x {}^{2} + \frac{1}{x {}^{2} } - 2 \times x \times \frac{1}{x} \\ = x {}^{2} + \frac{1}{x {}^{2} } - 2 \\ (x - \frac{1}{x} ) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } - 2 \\ =51 - 2 \\ = 49 \\ (x - \frac{1}{x}) {}^{2} = 49 \\ x - \frac{1}{x} = \sqrt{49} \\ x - \frac{1}{x} = 7$

Answer 2

$\Large{\bold{Hey\:\;\;Mate\;\;\:Here\:\:\:Is\:\;\;Your\;\;\:Answers}}$

$\underline{\bold{Given}}$

$\bold{{x}^{2} + \frac{1}{ {x}^{2} }}$

$\bold{And\: We \:Need \:To\: Find...}$

$\bold{{x} - \frac{1}{x}}$

So Now Let's Move For Solution ....

$\underline{\bold{Solution}}$

Now According To Question It's Given That ${x}^{2} + \frac{1}{ {X}^{2} } = 51$

Now To Find ...

$\bold{x - \frac{1}{x}}$

Follow The Simple Step ...

$\underline{\bold{Step-1)\:Find\:The\: Suitable\: Identity\:}}$

So The MosT Suitable Identity That Can Be Used Here Is...

$\boxed{\bold{{(a - b)}^{2} \:= {a}^{2} \:+ \:{b}^{2} -2ab}}$

$\underline{\bold{Step-2)\:Apply\: The \:Identity\: To \: Find\: Value \:Of\:\:{x}\:- \frac{1}{x} }}$

Now Applying This Identity ..

Where

$\underline{\bold{a\: = \:x \:\:And\: b \:= 1/x \:}}$

$\bold{ {(x\: - \:\frac{1}{x})}^{2} = \: {x}^{2} \:+ \:\frac{1}{ {x}^{2} } - 2(x) (\frac{1}{x} )}$

That Is ...

$\bold{{(x - \frac{1}{x})}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } \: - 2 \: (by \: \: \: cancelling \: x)}$

So Now In Question Its Given That 

$\bold{ {x}^{2} + \frac{1}{ {x}^{2} }}$

Now Putting This Value In the Obtained Equation We Have ...

$\bold{ {(x - \frac{1}{x}) }^{2} = 51 - 2}$

That Is ...

$\bold{ {(x - \frac{1}{x} )}^{2} = 49}$

Now Moving Square To RHS We Have 

$\bold{x - \frac{1}{x} = \sqrt{49}}$

That Is ...

$\bold{x - \frac{1}{x} = 7}$

So Now We Have Value Of 

$\boxed{\bold{x - \frac{1}{x} = 7}}$

$\underline{\bold{Hence\:The\: Required\:Answer\:Is\:....}}$

$\boxed{\boxed{\bold{7}}}$

$\Large{\bold{Thanks...}}$