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

1 Answer
Aug 9, 2017

Answer:

See a solution process below:

Explanation:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(1)# for #color(red)(a)#

#color(blue)(-4)# for #color(blue)(b)#

#color(green)(1)# for #color(green)(c)# gives:

#x = (-color(blue)((-4)) +- sqrt(color(blue)((-4))^2 - (4 * color(red)(1) * color(green)(1))))/(2 * color(red)(1))#

#x = (color(blue)(4) +- sqrt(color(blue)(16) - 4))/2#

#x = (color(blue)(4) +- sqrt(12))/2#

#x = (color(blue)(4) - sqrt(4 * 3))/2# and #x = (color(blue)(4) + sqrt(4 * 3))/2#

#x = (color(blue)(4) - sqrt(4)sqrt(3))/2# and #x = (color(blue)(4) + sqrt(4)sqrt(3))/2#

#x = (color(blue)(4) - 2sqrt(3))/2# and #x = (color(blue)(4) + 2sqrt(3))/2#

#x = color(blue)(4)/2 - (2sqrt(3))/2# and #x = color(blue)(4)/2 + (2sqrt(3))/2#

#x = 2 - sqrt(3)# and #x = 2 + sqrt(3)#