Question

If x2 + 1/x2 = 18, find the value of x -1/x

Answer 1

Given Equation is x^2 + 1/x^2 = 18

= > x^2 + (1/x^2) = 2 + 16

= > x^2 + (1/x^2) - 2 = 16

We can write it as,

= > x^2 + (1/x^2) - 2 * x * (1/x) = 16

We know that a^2 + b^2 - 2ab = (a - b)^2

= > (x - 1/x)^2 = 16

$= > Answer : \boxed { x - \frac{1}{x} = +4, -4 }}$

Hope it helps!

Answer 2

Answer:

Given $x^{2} +\frac{1}{x^{2} }$=18   ......   (1)

And we have to find $x-\frac{1}{x}$

Let $x-\frac{1}{x}$ be $y$

Squaring on both sides,

$(x-\frac{1}{x})^{2}$=$y^{2}$

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

From Eqn (1): 18-2=$y^{2}$

$y^{2}$=16, which means $y=\pm4$$= x-\frac{1}{x}$

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