How do you solve #x^2 + 5x - 2 = 0#?

1 Answer
Jan 5, 2017

Answer:

#x = -5/2+-sqrt(33)/2#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We can solve the given quadratic by completing the square and using the difference of squares identity with #a=(2x+5)# and #b=sqrt(33)#.

I will multiply it by #4# to start with, to cut down on the need for arithmetic involving fractions:

#0 = 4(x^2+5x-2)#

#color(white)(0) = 4x^2+20x-8#

#color(white)(0) = (2x)^2+2(2x)(5)+25-33#

#color(white)(0) = (2x+5)^2-(sqrt(33))^2#

#color(white)(0) = ((2x+5)-sqrt(33))((2x+5)+sqrt(33))#

#color(white)(0) = (2x+5-sqrt(33))(2x+5+sqrt(33))#

#color(white)(0) = 4(x+5/2-sqrt(33)/2)(x+5/2+sqrt(33)/2)#

Hence:

#x = -5/2+-sqrt(33)/2#