[Math] Why does completing the square work

completing-the-squarelinear algebra

I am currently learning about quadratics in high school and we've just done completing the square. Now I understand how to complete the square, I just don't understand why we can complete the square.

This is the generic complete the square formula

$ax^2 + bx + c = 0$

$x^2 + \frac{b}a x + \frac{c}a = 0$

$(x + \frac{b}{2a})^2 – \frac{b^2}{4a^2} + \frac{c}a = 0$

$(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} + \frac{c}a$

$x + \frac{b}{2a} = \frac{b}{2a} \pm \sqrt{\frac{c}a}$

$x = \frac{b}{2a} – \frac{b}{2a} \pm \sqrt{\frac{c}a}$

$x = \pm \sqrt{\frac{c}a}$

NB: I hope I got this right; I'm more used to numbers. If not, feel free to correct

And, as I find it easier to spot mistakes, a numerical example.

$2x^2 – 4x – 16 = 0$

$x^2 – 2x – 8 = 0$

$(x – 1)^2 – 1 – 8 = 0$

$(x – 1)^2 – 9 = 0$

$(x – 1)^2 = 9$

$x – 1 = \pm 3$

$x = 1 \pm 3$

What I don't understand is why line 3 works and how it can produce the correct answer. I don't see how we can divide it by 2 and then subtract its square.

My math teacher didn't really explain this as it's probably beyond what we have to learn but is there anyone who can tell me how this works?

Best Answer

"I just don't understand why we can complete the square."

The best way to see why it works, is to see a pictorial representation of what is going on, see:

Complete the square

The $x^2$ term is a square with both sides of length $x$. The $bx$ term is a rectangle with one side of length $x$ and one side of length $b$. When you divide by two you are splitting the $bx$ rectangle in two parts with still a side of length $x$ and the other side is now $\frac b2$.

When you place both rectangles as shown on the picture there is only a small square with sides of length $\frac b2$ missing. This small square has area $\frac {b^2}{4}$ and it is this part that you need to "complete the square".

If you have a term $a\ne 1$ in $ax^2+bx+c$ then you have to first "get it off" to get a square with both sides of length $x$. In that case the rectangle you get still has a side of length $x$ but its other side is now $\frac ba$ and this makes the area of the tiny missing square equal to $\frac {b^2}{4a}$.

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