Answer: [tex]x=\frac{(1-k^{2}) \pm (1+k^{2})}{2k}[/tex]
Step-by-step explanation:
We have the following quadratic equation:
[tex]x^{2} + (k-\frac{1}{k})x- 1=0[/tex]
Which is in the form: [tex]ax^{2} + bx + c=0[/tex]
And can be solved with the quadratic formula: [tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
Where [tex]a=1[/tex], [tex]b=k-\frac{1}{k}[/tex], [tex]c=1[/tex]
Knowing this, let's begin:
[tex]x=\frac{-(k-\frac{1}{k})\pm\sqrt{(k-\frac{1}{k})^{2}-4(1)(-1)}}{2(1)}[/tex]
[tex]x=\frac{-k+\frac{1}{k}\pm\sqrt{k^{2}-2k\frac{1}{k}+\frac{1}{k^{2}} +4}}{2}[/tex]
[tex]x=\frac{\frac{1-k^{2}}{k}\pm\sqrt{k^{2}+\frac{1}{k^{2}} +2}}{2}[/tex]
[tex]x=\frac{1-k^{2}}{2k} \pm \frac{\sqrt{\frac{k^{4}+1+2k^{2}}{k^{2}}}}{2}[/tex]
[tex]x=\frac{1-k^{2}}{2k} \pm \frac{\sqrt{k^{4}+2k^{2} +1}}{2k}[/tex]
Factoring [tex]k^{4}+2k^{2} +1[/tex] we have [tex](k^{2} +1 )^{2} [/tex]. hence:
[tex]x=\frac{1-k^{2}}{2k} \pm \frac{\sqrt{(k^{2}+1)^{2}}}{2k}[/tex]
[tex]x=\frac{(1-k^{2}) \pm (1+k^{2})}{2k}[/tex]
Solving for both options:
[tex]x_{1}=\frac{(1-k^{2}) + (1+k^{2})}{2k}=\frac{1}{k}[/tex]
[tex]x_{2}=\frac{(1-k^{2}) - (1+k^{2})}{2k}=-k[/tex]
