If \(x^2+\frac{1}{x^2}\) = 18, find the values of \(x+\frac{1}{x}\)

1.	If \(x^2+\frac{1}{x^2}\) = 18, find the values of \(x+\frac{1}{x}\) and \(x-\frac{1}{x}\).
Answer» x² + \(\frac{1}{x\times x}\) = 18 Adding 2 on both sides, we get x² + \(\frac{1}{x\times x}\) + 2 = 18 + 2 x² + \(\frac{1}{x\times x}\) + 2 × x × \(\frac{1}{x}\) = 20 (x + \(\frac{1}{x}\) )² = 20 x + \(\frac{1}{x}\) = \(2\sqrt{5}\) Given that, x² + \(\frac{1}{x\times x}\) = 18 Subtracting 2 from both sides, we get x² +\(\frac{1}{x\times x}\) - 2 × x × \(\frac{1}{x}\) = 18 – 2 (x - \(\frac{1}{x}\))² = 16 x - \(\frac{1}{x}\) = 4

If \(x^2+\frac{1}{x^2}\) = 18, find the values of \(x+\frac{1}{x}\) and \(x-\frac{1}{x}\).

Answer»

x² + \(\frac{1}{x\times x}\) = 18

Adding 2 on both sides, we get

x² + \(\frac{1}{x\times x}\) + 2 = 18 + 2

x² + \(\frac{1}{x\times x}\) + 2 × x × \(\frac{1}{x}\) = 20

(x + \(\frac{1}{x}\) )² = 20

x + \(\frac{1}{x}\) = \(2\sqrt{5}\)

Given that,

x² + \(\frac{1}{x\times x}\) = 18

Subtracting 2 from both sides, we get

x² +\(\frac{1}{x\times x}\) - 2 × x × \(\frac{1}{x}\) = 18 – 2

(x - \(\frac{1}{x}\))² = 16

x - \(\frac{1}{x}\) = 4