Question

if x+1/x=3,find x*2+1/x*2

Answer 1

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

___________ [GIVEN]

• We have to find the value of x² + $\dfrac{1}{ {x}^{2} }$

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

Squaring on both sides ..

$\implies$ $\bigg( {x \: + \: \dfrac{1}{x}\bigg)}^{2}$ = (3)²

We know that

(a + b)² = a² + b² + 2ab

Here.. a = x and b = $\dfrac{1}{x}$

$\implies$ x² + $\dfrac{1}{ {x}^{2} }$ + $2{ \cancel{x}} \: \times \: \dfrac{1}{ \cancel{x}}$ = 9

$\implies$ x² + $\dfrac{1}{ {x}^{2} }$ + 2 = 9

$\implies$ x² + $\dfrac{1}{ {x}^{2} }$ = 9 - 2

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$\implies$ x² + $\dfrac{1}{ {x}^{2} }$ = 7

______________ [ANSWER]

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Answer 2

Given the value of

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

To find the value of

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

•Here,I will be using the following identify:

(a+b)²=a²+2ab+b²

where a=x and b=1/x

Now,

$\sf{(x + \frac{1}{x}) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } + 2 } \\ \\ \implies \: \sf{x {}^{2} + \frac{1}{x {}^{2} } = 3 {}^{2} - 2 = 9 - 2 = 7 } \\ \\ \implies \: \boxed{ \sf{x {}^{2} + \frac{1}{x {}^{2} } = 7 }}$