How do you solve 2x^2 = 3x +92x2=3x+9 by completing the square?

1 Answer
Mar 6, 2017

See the entire solution process below:

Explanation:

First, subtract color(red)(3x)3x and color(blue)(9)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)2x23x9=3x+93x9

2x^2 - 3x - 9 = 3x - color(red)(3x) + 9 - color(blue)(9)2x23x9=3x3x+99

2x^2 - 3x - 9 = 0 + 02x23x9=0+0

2x^2 - 3x - 9 = 02x23x9=0

Next, factor the quadratic equation:

(2x + 3)(x - 3) = 0(2x+3)(x3)=0

Now, solve each term for 00:

Solution 1)

2x + 3 = 02x+3=0

2x + 3 - color(red)(3) = 0 - color(red)(3)2x+33=03

2x + 0 = -32x+0=3

2x = -32x=3

(2x)/color(red)(2) = -3/color(red)(2)2x2=32

(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