How do you solve #x^2-4x-1=0# using the quadratic formula?

1 Answer
Jul 1, 2016

Answer:

#x=(2-sqrt(5))# or #x=(2+sqrt(5))#

Explanation:

For the general quadratic equation:
#color(white)("XXX")color(red)(a)x^2+color(blue)(b)x+color(green)(c)=0#
the quadratic formula tells us that
#color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2(color(red)(a)))#

The given equation: #x^2-4x-1=0# can be written explicitly in the general form as:
#color(white)("XXX")x=color(red)(1)x^2+color(blue)(""(-4))+color(green)(""(-1))=0#

So the solution is
#color(white)("XXX")x=(-color(blue)(""(-4))+-sqrt(color(blue)(""(-4))^2-4color(red)(""(1))color(green)(""(-1))))/(2(color(red)(1))#

#color(white)("XXXX")=(4+-sqrt(20))/2#

#color(white)("XXXX")=2+-sqrt(5)#