How do you solve #2x^2 = 3x +9# by completing the square?

1 Answer
Mar 6, 2017

Answer:

See the entire solution process below:

Explanation:

First, subtract #color(red)(3x)# and #color(blue)(9)# from each side of the equation to put the quadratic equation in standard form:

#2x^2 - color(red)(3x) - color(blue)(9) = 3x + 9 - color(red)(3x) - color(blue)(9)#

#2x^2 - 3x - 9 = 3x - color(red)(3x) + 9 - color(blue)(9)#

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

#2x^2 - 3x - 9 = 0#

Next, factor the quadratic equation:

#(2x + 3)(x - 3) = 0#

Now, solve each term for #0#:

Solution 1)

#2x + 3 = 0#

#2x + 3 - color(red)(3) = 0 - color(red)(3)#

#2x + 0 = -3#

#2x = -3#

#(2x)/color(red)(2) = -3/color(red)(2)#

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

#x = -3/2#

Solution 2)

#x - 3 = 0#

#x - 3 + color(red)(3) = 0 + color(red)(3)#

#x - 0 = 3#

#x = 3#

#(2x)/color(red)(2) = -3/color(red)(2)#

The solution is: #x = -3/2# and #x = 3#