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

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 ${x}^{2} + 4 x - 7 = 0$
Step 1: ${x}^{2} + 4 x = 7$ (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 ${x}^{2} + 4 x + 4 = 7 + 4$
Step 4: Factor the trinomial: ${\left(x + 2\right)}^{2} = 11$
Step 5: Take the square root of both sides: $\sqrt{{\left(x + 2\right)}^{2}} = \sqrt{11}$
Step 6: $x + 2 = \pm \sqrt{11}$ be sure to use the $\pm$ on the square root!
Step 7: $x = - 2 \pm \sqrt{11}$ move the constant to the other side.

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

Solve ${x}^{2} - 8 x - 9 = 0$
${x}^{2} - 8 x = 9$
${x}^{2} - 8 x + 16 = 9 + 16$
${x}^{2} - 8 x + 16 = 25$
${\left(x - 4\right)}^{2} = 25$
$\sqrt{{\left(x - 4\right)}^{2}} = \sqrt{25}$
$x - 4 = \pm 5$
$x = 4 + 5 \mathmr{and} 4 - 5$
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.