How do you solve #y=5x^2+20x+23# using the completing square method?

1 Answer
Dec 27, 2016

Answer:

#x = -2+-sqrt((y-3)/5)#

Explanation:

I will assume that we want to solve the equation in the sense of expressing possible values of #x# in terms of #y#.

If you actually just wanted to know the zeros, we can then put #y=0# and evaluate the result.

Given:

#y = 5x^2+20x+23#

We will perform a sequence of steps to isolate #x# on one side.

First divide both sides by #5# to make the coefficient of #x^2# into #1#:

#y/5 = x^2+4x+23/5#

If we take half of the coefficient of #x# then we get the value #2#, so note that:

#(x+2)^2 = x^2+4x+4#

which matches the right hand side of our equation in the first two terms.

So we can proceed as follows:

#y/5 = x^2+4x+23/5#

#color(white)(y/5) = x^2+4x+4-4+23/5#

#color(white)(y/5) = (x+2)^2+3/5#

Subtracting #3/5# from both ends we get:

#(y-3)/5 = (x+2)^2#

Transposed:

#(x+2)^2 = (y-3)/5#

Take the square root of both sides, allowing for the possibility of either sign of square root to get:

#x+2 = +-sqrt((y-3)/5)#

Finally subtract #2# from both sides to find:

#x = -2+-sqrt((y-3)/5)#

If we put #y=0# to find the zero of the quadratic then we get a pair of non-Real Complex values:

#x = -2+-sqrt((0-3)/5) = -2+-sqrt(-3/5) = -2+-sqrt(3/5)i = -2+-sqrt(15)/5i#