Question

if x-1/x=3 then what is the value of x²+1/x²​

Answer 1

Answer:

x-1/x=3 (a-b)^2=a^2+b^2-2ab

(x-1/x)^2=x^2+1/x^2-2(x*1/x)

(3)^2=x^2+1/x^2-2

9+2 =x^2+1/x^2

x^2+1/x^2=11

Answer 2

Understanding the concept:

•We are given that x -1/x = 3 and we are said to find the value of x^2 +1/x^2 so first of all as we know that the formula of a^2 +b^2 = (a-b)^2+2ab so we are provided the formula then we will put the values which are given then find the answer.So let's find!

Given:-

•x -$\sf{\dfrac{1}{x}}$ = 3

To Find:-

•$\sf{x}^{2}$+$\sf{\dfrac{1}{{x}^{2}}}$

Solution:-

Given,

• x - $\sf{\dfrac{1}{x}}$ = 3

As we know,

$\: \sf \large \: ( {a}^{2} + {b}^{2} ) = {(a - b)}^{2} + 2 ab$

Now put the value

$\: \sf \implies \: ( {x}^{2} + \frac{ 1 }{ {x}^{2} } ) = {(x - \frac{1}{x}) }^{2} + 2 \cancel x. \frac{1}{ \cancel x} \\ \\ \: \sf \implies \: {3}^{2} + 2 \\ \\ \: \sf \implies \: 9 + 2 \\ \\ \: \sf \implies \: 11$

$\: \sf \therefore \: the \: value \: of \: ( {x}^{2} + \frac{1}{ {x}^{2} } ) \: is \: 11.$

Additional Information:-

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

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