Question

Can you justify if completing the square is a good method for solving when the Discriminant is negative?
Use any of these three functions as an example
f(x)=x^2-4
g(x)=x^2-2
h(x)=-3x^2 +11i

Answer 1

completing the square is the proof of the quadratic formula.

Answer 2

\[ax^2+bx+c=0\]
\[a(x^2+\frac{b}{a}x)+c=0\]
\[a(x^2+\frac{b}{a}x\pm \frac{b^2}{4a^2})+c=0\]
\[a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}+c=0\]
\[a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c=0\]
\[a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c\]
\[(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac ca\]
\[(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac {4ac}{4aa}\]
\[x+\frac{b}{2a}=\pm\sqrt{\frac{b^2}{4a^2}-\frac {4ac}{4a^2}}\]
\[x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]

Answer 3

what?

Answer 4

like i said: completing the square is the proof of the quadratic formula.

since the quadratic formula provides solutions; then so does completing the square ... they are 2 sides of the same coin :/

Answer 5

do you know how to convert f(x) into general vertex form?