Question

if X square +1/x square =18 find the value of x ;x+1/x and x-1/x​

Answer 1

Given that $x^2+\frac{1}{x^2} =18$

Let's find the value of $x+\frac{1}{x}$ and $x-\frac{1}{x}$ first.

Know that :

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

We are given that $x^2+\frac{1}{x^2} =18$, so substitue that in.

You must be really careful about the sign( - or + ) of x.

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

∴ $x+\frac{1}{x} =\pm 2\sqrt{5}$

$(x-\frac{1}{x} )^2=18-2$

∴ $x-\frac{1}{x} = \pm 4$

Find the value of x.

Given that : $x^2+\frac{1}{x^2} =18$

We can multiply x² to both sides.

$x^4-18x^2+1=0$

Does it look complicated? Yes but it can be factorized by identities.

First, divide the term which is degree 2.

$(x^4-2x^2+1)-16x^2=0$

$(x^2-1)^2-(4x)^2=0$

Second, apply the identity $a^2-b^2=(a+b)(a-b)$.

$(x^2+4x-1)(x^2-4x-1)=0$

∴ $x^2+4x-1=0$ or $x^2-4x-1=0$

Last, apply quadratic formula.

∴ $x=-2\pm \sqrt{5}$ or $x=2\pm \sqrt{5}$

From above, x can be both positive or negative.

So, $x+\frac{1}{x} =\pm2\sqrt{5}$ and $x-\frac{1}{x} = \pm 4$ are correct.

I am happy to help you. Thank you!