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

Question

Kennedy Adams · Accepted Answer

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#

Hazel Anglin · Answer

undefined To solve the equation x^2 + 3x - 8 = 0 by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   x^2 + 3x = 8

2. To complete the square, take half of the coefficient of x (which is 3/2) and square it (which is 9/4):
   Add and subtract 9/4 inside the parentheses:
   x^2 + 3x + 9/4 - 9/4 = 8

3. Rewrite the left side as a square:
   (x + 3/2)^2 = 8 + 9/4

4. Simplify the right side:
   (x + 3/2)^2 = 32/4 + 9/4
   (x + 3/2)^2 = 41/4

5. Take the square root of both sides:
   x + 3/2 = ±√(41/4)

6. Simplify the square root:
   x + 3/2 = ±√41 / 2

7. Solve for x:
   x = -3/2 ± √41 / 2