Question

if (x+1/x)=4, find the value of (x^2+1/x^2)​

Answer 1

Given:-

$x + \frac{1}{x} = 4$

To find:-

$The \: value \: of \: \: {x}^{2} + \frac{1}{ {x}^{2} }$

Solution:-

Let's solve the problem

we have

$x + \frac{1}{x} = 4$

Squaring on both sides, we get

$\bigg(x + \frac{1}{x} \bigg)^{2} = (4 {)}^{2}$

✯Using algebraic Identity

(a+b)² = a² + 2ab + b²

Now,

$⟹ {x}^{2} + 2 \times x \times \frac{1}{x} + \bigg( \frac{1}{ {x}^{2} } \bigg)^{2} = 16$

$⟹ {x}^{2} + 2 \times \cancel{x} \times \frac{1}{ \cancel{x}} + \frac{1}{ {x}^{2} } = 16$

$⟹ {x}^{2} + 2 + \frac{1}{ {x}^{2} } = 16$

$⟹ {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2$

$⟹ {x}^{2} + \frac{1}{ {x}^{2} } = 14$

Answer:-

$Hence, the \: value \: of \: {x}^{2} + \frac{1}{ {x}^{2} } \: is \: 14$

• I hope it's help you...☺