How do you find the roots, real and imaginary, of #y= x^2 + 4x - 3 # using the quadratic formula?

1 Answer
Feb 29, 2016

Answer:

See explanation...

Explanation:

#x^2+4x-3# is of the form #ax^2+bx+c# with #a=1#, #b=4# and #c=-3#.

This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-4+-sqrt((-4)^2-(4*1*(-3))))/(2*1)#

#=(-4+-sqrt(16+12))/2#

#=(-4+-sqrt(28))/2#

#=(-4+-sqrt(2^2*7))/2#

#=(-4+-2sqrt(7))/2#

#=-2+-sqrt(7)#

Alternative Method

The difference of squares identity can be written:

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

Complete the square and use this with #a=x+2# and #b=sqrt(7)# as follows:

#x^2+4x-3#

#=x^2+4x+4-7#

#=(x+2)^2-(sqrt(7))^2#

#=((x+2)-sqrt(7))((x+2)+sqrt(7))#

#=(x+2-sqrt(7))(x+2+sqrt(7))#

Hence zeros: #x=-2+-sqrt(7)#