Question

if x+1/x=3 find the value of x square +1/x square

Answer 1

$\textsf{hey\: mate!\: here's\: your\: answer!}$
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since, given :-

x + $\frac{1}{x}$ = 3

so, by the identity :-

(a+b)² = a² + b² + 2ab

we will solve ,

=> ( x + $\frac{1}{x}$ )² = (x)² + ( $\frac{1}{x}$ )² + 2(x)($\frac{1}{x}$ )

now, we will substitute the value of x + $\frac{1}{x}$ ,

=> (3)² = x² + $\frac{1}{{x}}^{2}$ + 2

then, we will take 2 to other side, and 2 is positive, so taking it to other side, it will become -2.

=> 9 - 2 = x² + $\frac{1}{{x}}^{2}$

=> x² + $\frac{1}{{x}}^{2}$ = 7 .....answer.

Answer 2

Here's your answer!!

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It's given that,

$= > x + \frac{1}{x} = 3$

We have to find value of ,

${x}^{2} + \frac{1}{ {x}^{2} }$
So,

$= > x + \frac{1}{x} = 3$

By squaring both side ,we get :-

$= > {(x + \frac{1}{x}) }^{2} = {(3)}^{2}$

We will use one identity,that is ,

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$= > {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$
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$= > {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 9$

$x \: get \: cancelled \: from \: both \: side$


$= > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 9$

$= > {x}^{2} + \frac{1}{{x}^{2} } = 7$
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Hope it helps you!! :)