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 x2, and an even number for the coefficient on the linear term (the x term).

Example: Solve x2+4x7=0
Step 1: x2+4x=7 (move the constant to the opposite side)
Step 2: take half of the "4", and square that number. 22=4
Step 3: Add that number to both sides x2+4x+4=7+4
Step 4: Factor the trinomial: (x+2)2=11
Step 5: Take the square root of both sides: (x+2)2=11
Step 6: x+2=±11 be sure to use the ± on the square root!
Step 7: x=2±11 move the constant to the other side.

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

Solve x28x9=0
x28x=9
x28x+16=9+16
x28x+16=25
(x4)2=25
(x4)2=25
x4=±5
x=4+5or45
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.