How do you solve the quadratic equation by completing the square: #3x^2+9x-1=0#?

1 Answer
Jul 16, 2015

Answer:

#x = -3/2 +-sqrt(31)/(2sqrt(3))#
#color(white)("XXXX")##color(white)("XXXX")##color(white)("XXXX")#(by completing the square)

Explanation:

Given #3x^2+9x-1=0#

Move the constant to the right side and extract the common factor of #3# from the left side:
#color(white)("XXXX")##3(x^2+3x) = 1#

If #x^2+3x# are the first two terms of the expansion of a squared binomial:
#color(white)("XXXX")##(x+a)^2 = (x^2+2ax+a^2)#

then #a= 3/2# and #a^2 = (3/2)^2 = 9/4#
#color(white)("XXXX")#which is the amount we will need to add (inside the parentheses) to "complete the square)

#color(white)("XXXX")#Note that adding #9/4# inside the parentheses is the same as adding #3*(9/4)# and this amount will need to be added to the right side as well to keep the equation balanced.

#color(white)("XXXX")##3(x^2+3x+(3/2)^2) = 1 + 27/4#

Rewriting as a squared binomial (and simplifying the right side)
#color(white)("XXXX")##3(x+3/2)^2 =31/4#

Dividing both sides by #3#
#color(white)("XXXX")##(x+3/2)^2 = 31/12#

Taking the square root of both sides:
#color(white)("XXXX")##x+3/2 = +- sqrt(31/12)#

Subtracting #3# from both sides
#color(white)("XXXX")##x = -3/2 +-sqrt(31)/(2sqrt(3))#