Question

If x + 1 by x equal to 3 then find the value of x square + 1 by x square

Answer 1

$\underline{\mathfrak{\huge{The\:Question:}}}$

If the value of $x + \frac{1}{x}$ = 3, find the value of $x^{2} + \frac{1}{x^{2}}$.

$\underline{\mathfrak{\huge{Here's\:Your\:Answer:}}}$

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

Do the squaring on both the sides :-

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

Now, use the identity, which is :-

$(a+b)^{2} = a^{2} + b^{2} + 2ab$

In the last equation formed. Now, next , solve it :-

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

Now, you need to solve it further :-

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

Take the 2 from Left hand side to the Right Hand Side and then subtract it from 9 :-

$x^{2} + \frac{1}{x^{2}}$ = 7

There's your answer !

Answer 2

Question : If x + 1/x = 3 , then find the value of x^2 + ( 1/x^2).

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

Let's square the both sides of the equation. We get :

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

Now, We can easily simplify the R. H. S , In the L. H. S , use the identity :

$\tt{(a + b) {}^{2} = a {}^{2} + 2ab + {b}}^{2}$

$\tt{\therefore x {}^{2} + 2 \times x \times \frac{1}{x} \times( \frac{1}{x} ) {}^{2} = 9}$

Simplifying step by step ,

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

Taking 2 to the R.H.S

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

Simplifying the equation by cancelling x in L. H. S

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

Thus, We got the answer.

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