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

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.