How do you solve the quadratic equation by completing the square: #4x^2 + 9 = 12x#?

1 Answer
Jul 15, 2015

Answer:

#x=3/2# is a "double root" (a root of multiplicity 2 ).

Explanation:

First, rearrange so the #x^2# and #x# terms are on the left and the constant term is on the right: #4x^2-12x=-9#.

Next, factor out the coefficient of #x^2# on the left: #4(x^2-3x)=-9#.

Next, you can either divide both sides by 4 before completing the square or complete the square first. Let's complete the square on #x^2-3x# first. Take the coefficient of #x#, which is #-3#, divide it by 2 to get #-3/2#, then square this number to get #9/4#. Add that number inside the parentheses and balance on the right side of the equation to get:

#4(x^2-3x+9/4)=-9+4*9/4=0# (it doesn't usually come out to be zero on the right here, that's just a coincidence).

Once you've done this procedure, the expression inside the parentheses will be a perfect square. In this case, we can write:

#4(x-3/2)^2=0#

In general, we can now divide both sides by the coefficient in front, which is 4 in this case. Since #0/4=0#, we get #(x-3/2)^2=0#.

Now take the #\pm# square root of both sides. Since #\pm sqrt(0)=0#, there is only one equation and one root of the original equation: #x-3/2=0# so #x=3/2# is a root of multiplicity 2.