Question

if x+1/x=3,find x^2+1/x^2​

Answer 1

Given that,

    x + 1/x = 3

We have to find,

    x² + 1/x²

So, step - by - step procedure is given below.

→  Square both the sides of the given equation.

    (x + 1/x)² = 3²

⇒  x²  +  2 · x · 1/x  +  (1/x)²  =  9

⇒  x² + 2 + 1/x²  =  9

→  Subtract 2 from both sides of the equation obtained.

    x² + 2 + 1/x² - 2  =  9 - 2

⇒  x² + 1/x²  =  7

Thus we get the answer!

It is none other than 7.

For such questions, we can use a simple formula derived from this procedure.

$\large\textsf{If\ \ x+\dfrac{1}{x}=k, \ \ then\ \ x^2+\dfrac{1}{x^2}=k^2-2. }$

So, here,  k = 3.

Hence the answer will be  k² - 2 = 3² - 2 = 9 - 2 = 7.

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\large{\boxed{ \tt {x}^{2} + \dfrac{1}{ {x}^{2} } = 7}}$

$\mathfrak{\large{\underline{\underline{Explanation-}}}}$

Given :- $\sf x + \dfrac{1}{x} = 3$

To find :- $\sf x^2 + \dfrac{1}{x^2}$

Solution :-

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

Squaring on both sides

$\tt (x + \dfrac{1}{x})^{2} = {(3)}^{2}$

We know that (x + y)² = x² + y² + 2xy

Here x = x, y = 1/x

By substituting the values in the identity we have

$\tt {x}^{2} + (\dfrac{1}{x})^{2} + 2(x)( \dfrac{1}{x}) = 9$

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

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

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

$\Huge{\boxed{ \tt {x}^{2} + \dfrac{1}{ {x}^{2} } = 7}}$