Question

give me the answer it's urgent ​

Answer 1

Given

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

To Find

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

Answer and explanation

We know the identity

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

So let's use it here

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

We get

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

Since x and 1/x cancel out

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

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

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