How do you solve by completing the square: #ax^2+bx+c=0#?

1 Answer
Apr 1, 2015

#ax^2 + bx +c = 0#

Divide all terms by #a# so as to reduce the coefficient of #x^2# to #1#
#x^2 +b/a x + c/a=0#

Subtract the constant term from both sides of the equation
#x^2+b/a x = - c/a#

To have a square on the left side the third term (constant) should be
#(b/(2a))^2#

So add that amount to both sides
#x^2 +b/a x +(b/(2a))^2 = (b/(2a))^2 - c/a#

Re-write the left-side as a square
#(x+(b/(2a)))^2 = (b/(2a))^2 -c/a#

Take the square root of both sides (remembering that the result could be plus or minus)
#x+(b/(2a)) = +-sqrt((b/(2a))^2 -c/a)#

Subtract the constant term on the left side from both sides
#x = +-sqrt((b/(2a))^2 -c/a) - (b/(2a))#

or, with some simplification

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

(the standard form for solving a quadratic)