How do you complete the square when a quadratic equation has a coefficient?

1 Answer
Feb 20, 2015

It is easier to explain with an example:

#3x^2+7x-5=0rArr3(x^2+7/3x)-5=0#

Now, since #7/3x# will have to be the double product of the first and the second terms of the square, if we divide it for the double of the first (#2x#) we can obtain the second:

#(7/3x)/(2x)=7/6#, so:

#3(x^2+7/3x+(7/6)^2-(7/6)^2)-5=0rArr#

#3(x^2+7/3x+49/36-49/36)-5=0rArr#

#3(x^2+7/3x+49/36)-3*49/36-5=0rArr#

#3(x+7/6)^2-49/12-5=0rArr#

#3(x+7/6)^2=(49+60)/12rArr#

#3(x+7/6)^2=109/12rArr# and, if you want to solve:

#(x+7/6)^2=109/36rArr(x+7/6)=+-sqrt(109/36)rArr#

#x=-7/6+-sqrt109/6=(-7+-sqrt109)/6#.