Question

In the quadratic equation \[x^2 + \left(k-\frac{1}{k}\right)x - 1 = 0,\]solve for $x$ in terms of $k$.

Answer 1

Answer:
`x=((1-k^(2)) \pm (1+k^(2)))/(2k)`

Explanation:

We have the following quadratic equation:

`x^(2) + (k-(1)/(k))x- 1=0`

Which is in the form:
`ax^(2) + bx + c=0`

And can be solved with the quadratic formula:
`x=\frac{-b\pm\sqrt{b^(2)-4ac}}{2a}`

Where
`a=1`,
`b=k-(1)/(k)`,
`c=1`

Knowing this, let's begin:

`x=\frac{-(k-(1)/(k))\pm\sqrt{(k-(1)/(k))^(2)-4(1)(-1)}}{2(1)}`

`x=\frac{-k+(1)/(k)\pm\sqrt{k^(2)-2k(1)/(k)+(1)/(k^(2)) +4}}{2}`

`x=\frac{(1-k^(2))/(k)\pm\sqrt{k^(2)+(1)/(k^(2)) +2}}{2}`

`x=(1-k^(2))/(2k) \pm \frac{\sqrt{(k^(4)+1+2k^(2))/(k^(2))}}{2}`

`x=(1-k^(2))/(2k) \pm \frac{\sqrt{k^(4)+2k^(2) +1}}{2k}`

Factoring
`k^(4)+2k^(2) +1` we have
`(k^(2) +1 )^(2)`. hence:

`x=(1-k^(2))/(2k) \pm \frac{\sqrt{(k^(2)+1)^(2)}}{2k}`

`x=((1-k^(2)) \pm (1+k^(2)))/(2k)`

Solving for both options:

`x_(1)=((1-k^(2)) + (1+k^(2)))/(2k)=(1)/(k)`

`x_(2)=((1-k^(2)) - (1+k^(2)))/(2k)=-k`