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

1 Answer
Jul 28, 2015

Answer:

#x_(1,2) = 3/2 +- sqrt(69)/6#

Explanation:

Starting from the general form of the quadratic equation

#ax^2 + bx + c = 0#

divide verything by #a# to get it to the form

#x^2 + b/ax = -c/a#

In your cae, #a=3#, which means that you get

#(color(red)(cancel(color(black)(3))) x^2)/color(red)(cancel(color(black)(3))) - 9/3x = -1/3#

#x^2 - 3x = -1/3#

To solve this quadratic by completing the square, you need to write the left side of the equation as the square of a binomial. This can be done by adding a term to both sides of the equation.

The coefficient of the #x#-term will help you determine what you need to add to both sides.

So, divide this coefficient by 2, then square the result to get

#(-3)/2 = -3/2#, then

#(-3/2)^2 = 9/4#

Adding #9/4# to both sides of the equation will produce

#x^2 - 3x + 9/4 = -1/3 + 9/4 = 23/12#

The general formula for the square of a binomial looks like this

#color(blue)((x + n)^2 = x^2 + 2n + n^2)#

You can use this formula to rewrite the left side of the equation as

#x^2 - 3x + 9/4 = x^2 + 2 * (-3/2)x + (-3/2)^2#

#x^2 - 3x + 9/4 = (x-3/2)^2#

Your quadratic becomes

#(x-3/2)^2 = 23/12#

Take the square root of both sides to get

#sqrt((x-3/2)^2) = sqrt(23/12)#

#x-3/2 = +-sqrt(23)/sqrt(12) = +- (sqrt(23) * sqrt(3))/(2 * 3) = +- sqrt(69)/6#

The two solutions to your quadratic equation will thus be

#x_1 = color(green)(3/2 + sqrt(69)/6)# and #x_2 = color(green)(3/2 - sqrt(69)/6)#