How do you find the roots, real and imaginary, of #y= (x+2)^2-x-5 # using the quadratic formula?

1 Answer
Jul 18, 2017

Answer:

See a solution process below:

Explanation:

First, we need to expand the squared term on the right side of the equation using the rule:

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

Substituting #x# for #a# and #2# for #b# gives:

#y = (x + 2)^2 - x - 5#

#y = x^2 + (2 xx x xx 2) + 2^2 - x - 5#

#y = x^2 + 4x + 4 - x - 5#

We can now group and combine like terms to put the equation in standard form for a quadratic.

#y = x^2 + 4x - x + 4 - 5#

#y = x^2 + 4x - 1x + 4 - 5#

#y = x^2 + (4 - 1)x + (4 - 5)#

#y = x^2 + 3x + (-1)#

#y = x^2 + 3x - 1#

The quadratic formula states:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

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

Substituting #1# for #a#; #3# for #b# and #-1# for #c# gives:

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

#x = (-3 +- sqrt(9 - (-4)))/2#

#x = (-3 +- sqrt(9 + 4))/2#

#x = (-3 +- sqrt(13))/2#

Or

#x = -3/2 + sqrt(13)/2# and #x = -3/2 - sqrt(13)/2#