Question

if x minus one by x is equal to 9 find the value of x square + 1 by x square​

Answer 1

$\red {\huge {\underline{ \frak{Your \: answEr : }}}}$

$\large\rightarrow \tt{{x} - \frac{1}{x} = 9}$

$\blacksquare \: \frak{squaring \: both \: sides}$

$\large\rightarrow \tt{ {(x - \frac{1}{x} )}^{2} = {(9)}^{2} }$

$\large\rightarrow \tt{ {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times x \times \frac{1}{x} = 81}$

$\large\rightarrow \tt{ {x}^{2} + \frac{1}{ {x}^{2} } = 81 + 2 }$

$\pink{\huge\rightarrow \boxed{\tt{ {x}^{2} + \frac{1}{ {x}^{2} } = 83 }}}$

$\huge{\red{\ddot{\smile}}}$

Answer 2

Given: Given expression is

$x - \frac{1}{x} = 9$

To find: We have to find

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

Solution:

To determine the value of $x^2+\dfrac{1}{x^2}$ we have to follow the following steps as follows-

The expression $x^2+\dfrac{1}{x^2}$ can be written as-

${(x - \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} \\ {(x - \frac{1}{x} )}^{2} + 2$

Now we know the value of x-1/x.

The given value of x-1/x is 9.

Putting these values in the above expression we get-

${9}^{2} + 2 \\ = 81 + 2 \\ = 83$

Thus the simplified value of the expression is 83.

The value of $x^2+\dfrac{1}{x^2}$ is 83.