How do you solve x^2-4x-1 >0?

May 14, 2018

$x > 2 \pm \sqrt{5}$

Explanation:

${x}^{2} - 4 x - 1 > 0$

Factorise the left into the form ${\left(x + b\right)}^{2} + c > 0$
Which becomes ${\left(x - 2\right)}^{2} - 5 > 0$

Then move the 5 over to the other side.
Which becomes ${\left(x - 2\right)}^{2} > 5$

Then Square root both sides.
Which becomes $x - 2 > \sqrt{5}$

Then move the 2 over ($+ 2$)
Which gives you your answer $x > 2 + \sqrt{5}$

$x > 2 \pm \sqrt{5}$