How do you solve by completing the square #x^2+3x-8=0#?

1 Answer
Apr 3, 2015

Solving a quadratic expression by completing the square means to manipulate the expression in order to write it in the form
#(x+a)^2=b#
So, if #b\ge 0#, you can take the square root at both sides to get
#x+a=\pm\sqrt{b}#
and conclude #x=\pm\sqrt{b}-a#.

Now, we have #(x+a)^2=x^2+2ax+a^2#. Since you equation starts with #x^2+3x#, this means that #2ax=3x#, and so #a=3/2#.
Adding #41/4# at both sides, we have
#x^2+3x+9/4=41/4#
Which is the form we wanted, because now we have
#(x+3/2)^2=41/4#
Which leads us to
#x+3/2=\pm\sqrt{41/4}=\pm \sqrt{41}/2# and finally #x=\pm\sqrt{41}/2-3/2#