I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?)

Solution 1:

Every quadratic can have its square completed, which is in fact where the quadratic formula comes from.

The method I use is to factor out the leading coefficient,

$$ \begin{align} \color{red}{a} x ^2 + bx +c &= \color{red}{a}\left( x^2 + \frac{b}{\color{red}{a}} x \right) + c \\ &= a\left( \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} \right) + c \\ &= a\left(x+\frac{b}{2a}\right)^2 + c-\frac{b^2}{4a}, \end{align} $$

if you solve for when this is equal to zero you will get the quadratic formula.

Solution 2:

The proposed way of completing the square is correct. The identity is: $$ ax^2 + bx + c = \left(\sqrt{a}x + \frac{b}{2 \sqrt{a}}\right)^2 + \left(c-\frac{b^2}{2a}\right). $$

One can of course also do what is more usually done and write $$ ax^2+bx+c = a\left( x + \frac{b}{2a} \right)^2 + \left(c-\frac{b^2}{2a}\right). $$ The latter way is done for the sake of convenience, not because it is the only way possible.

PS: Someone points out in comments that this works only if $a>0$. If $a<0$, one can write: $$ ax^2 + bx + c = -\left(\sqrt{|a|}x + \frac{b}{2 \sqrt{|a|}}\right)^2 + \left(c-\frac{b^2}{2a}\right). $$

Solution 3:

It seems like your questions have been answered, so this response is more of an addition. Completing the square is a useful rearrangement in and of itself - beyond the usage for solving quadratics. You may be aware of its application in placing the equation of a conic section in standard form.

For example, consider

$$x^2 + 2x + 2y^2 + y + 3 = 6$$

This is an ellipse, but our equation does not tell us much information. Rearranging into standard form will reveal quite a bit of info. Standard form for a horizontal ellipse is given as

$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$

where $(h,k)$ is the center and $a$ and $b$ represent the semi-major and semi-minor axes respectively (the longest and shortest "radius" of this "eccentric circle").

In order to place our ellipse into standard form, we complete the square twice:

$$x^2 + 2x + 2y^2 + 4y + 3 = 6$$ $$\left(x^2 + 2x \color{red}{+1}\right) + 2\left(y^2 + 4y \color{red}{+4}\right)+ 3 = 6 \color{red}{+1} \color{red}{+8} $$ $$(x+1)^2+ 2(y+2)^2+ 3 = 15 $$ $$(x+1)^2+ 2(y+2)^2 = 12 $$ $$\frac{(x+1)^2}{12}+ \frac{(y+2)^2}{6} = 1 $$

We now see that we have an ellipse centered at $(h,k) = (-1,-2)$ with semi-major axis $a = \sqrt{12}$ and semi-minor axis $b=\sqrt6$. This matches the graph.

Solution 4:

If $a$ is not a “perfect square” there's no problem either. If your equation is $$ ax^2+bx+c=0 $$ then it's equivalent to $$ 4a^2x^2+4abx+4ac=0 $$ and completing the square is more evident: $$ 4a^2x^2+4abx+b^2-b^2+4ac=0 $$ or $$ (2ax+b)^2-(b^2-4ac)=0 $$ If you just want to factor the polynomial $ax^2+bx+c$ (with $a\ne0$, of course, just do the same: $$ ax^2+bx+c=\frac{1}{4a}(4a^2x^2+4abx+4ac)= \frac{1}{4a}\bigl((2ax+b)^2-(b^2-4ac)\bigr) $$ If $b^2-4ac<0$ there's nothing else to do, because the polynomial is irreducible over the reals; if $b^2-4ac=0$ it is $$ \frac{1}{4a}(2ax+b)^2 $$ and, if $b^2-4ac>0$ you get $$ ax^2+bx+c=\frac{1}{4a}(x+2a-\sqrt{b^2-4ac})(x+2a+\sqrt{b^2-4ac}) $$