How do you solve the equation #x^2+3x-18=0# by completing the square?

1 Answer
Jan 8, 2017

Answer:

#x = 3" "# or #" "x=-6#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(2x+3)# and #b=9# as follows:

#0 = 4(x^2+3x-18)#

#color(white)(0) = 4x^2+12x-72#

#color(white)(0) = (2x)^2+2(2x)(3)+9-81#

#color(white)(0) = (2x+3)^2-9^2#

#color(white)(0) = ((2x+3)-9)((2x+3)+9)#

#color(white)(0) = (2x-6)(2x+12)#

#color(white)(0) = (2(x-3))(2(x+6))#

#color(white)(0) = 4(x-3)(x+6)#

Hence:

#x = 3" "# or #" "x=-6#