Can every quadratic be solved by using the completing the square method?

1 Answer
Oct 23, 2014

Can every quadratic be solved by using the completing the square method? Yes, it sure seems so.....but, I wouldn't want to do it that way every time!!

Certainly, every quadratic can be solved by the quadratic formula, but I also wouldn't want to use it every time either.

I think you should develop some strategies for which method is best in which circumstances. Purple Math link

There are even websites that will solve the problems for you: Solver

For completing the square, I look for two things: "1" as the lead coefficient (on the #x^2#, and an even number for the coefficient on the linear term (the x term).

Example: Solve #x^2 + 4x - 7 = 0#
Step 1: #x^2 + 4x = 7# (move the constant to the opposite side)
Step 2: take half of the "4", and square that number. #2^2=4#
Step 3: Add that number to both sides #x^2+4x +4= 7 + 4#
Step 4: Factor the trinomial: #(x+2)^2= 11#
Step 5: Take the square root of both sides: #sqrt((x+2)^2)=sqrt(11)#
Step 6: #x+2= +-sqrt(11)# be sure to use the #+-# on the square root!
Step 7: #x=-2+-sqrt(11)# move the constant to the other side.

Phew, that's a lot of steps! And it definitely takes practice. One more example:

Solve #x^2-8x - 9 = 0#
#x^2-8x = 9#
#x^2-8x+16=9+16#
#x^2-8x+16=25#
#(x-4)^2=25#
#sqrt((x-4)^2)=sqrt(25)#
#x-4=+-5#
#x=4+5 or 4-5#
so, x = 9 or -1. Wow!

Of course, those nice, rational answers tell me that the original problem could have been solved by factoring: (x-9)(x+1)=0
Use the zero product property to solve x = 9 or x = -1.