How do you complete the square for negative quadratics I.e. in the form -ax2 bx c?

Answer 1

Without using complex numbers, which is a bad idea, you can't. But you can do something which is almost as good: instead of dealing with -ax^2+bx+c, you can take -(ax^2-bx-c), and then complete the square for the inside quadratic. This will give you something like -((rx+s)^2+t), which is usually good enough. For example, if you're completing the square in order to solve a quadratic equation, the sign on the outside doesn't matter: if -((rx+s)^2+t)=0 then (rx+s)^2+t=0 as well (because the negative of 0 is 0). A different approach, which is usually more useful, is to factor out the a, regardless of whether it's positive or negative. Then instead of completing the square for ax^2+bx+c, you complete the square for x^2+b/ax+c/a, and multiply the final result by a.