How do you solve #x^2+3x+21=22# by completing the square?

2 Answers
Mar 28, 2017

Answer:

#x = -3/2 +- sqrt(13)/2#

Explanation:

Steps:

Subtract 21 from both sides

#x^2+3x + ? = 1#

In order to find ?, you must divide #3# by #2# or #(b/2)#and square it and add it to both sides:

#? = (3/2)^2 =9/4#

#x^2+3x + 9/4 = 1 +9/4#

Now we can factor the left side and simplify the right.

#(x+3/2)^2 = 4/4+9/4#

#(x+3/2)^2 = 13/4#

Finally, solve for #x#

#x+3/2=0 -> x=-3/2#

#x^2-3/14=0 -> x^2 = 13/14 -> x = +- sqrt(13/4) -> x = +- sqrt(13)/2#

Mar 28, 2017

Answer:

# x = (-3 +- sqrt13)/2#

Explanation:

Squares of form #(x+a)^2# are expanded like this:

#x^2 + 2ax + a^2#

Since our equation starts with #x^2 + 3x...#, we know that #3 = 2a#, so #3/2 = a#. That means that to complete the square, we need the last term to be #a^2 = (3/2)^2 = 9/4#.

To do this, we can subtract #18.75# from both sides and simplify.

#color(white)"X"x^2 + 3x + 21 = 22#

#color(white)"X" x^2 + 3x + 9/4 = 13/4#

#color(white)"XX.." (x + 1.5)^2 = 13/4#

Now we can take the square root of both sides. Remember that the square root can be either positive or negative, since when squared, the negative sign disappears.

#color(white)"" x + 3/2 = +- sqrt13/2#

#color(white)"XXX." x = (-3 +- sqrt13)/2#

Final Answer