Why does completing the square work?

Question

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?

Answer 1

"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:

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 $\frac {b^2}{4a}$.

Answer 2

The goal is to get on one side of the equation a term that is a square. Then you take the square root on both sides of the equation (noting that there is a positive and a negative solution) to achieve the final solution.

Answer 3

The third line is wrong because you have added $\frac {b^2}{4a^2}$ and subtracted $\frac {b^2}{a^2}$. When you go from the fourth line to the fifth line you are not taking the square root correctly. You lost a $\pm$ sign on the left and the right is not the square root of the line above. You seem to be thinking $\sqrt{x-y}=\sqrt x \pm \sqrt y$, which is not true.

Answer 4

$ax^2 + bx + c = 0$

$x^2 + \frac ba x + \frac c a= 0$

Now we notice that $x^2 + \frac ba x +....$ looks like the start of $x^2 + 2kx + k^2 = (x+k)^2$. Can we get it into that format by adding and subtraction things? We ne $x^2 + 2kx + k^2 + ...stuff ... = x^2 + \frac ba x + .... different stuff...$ So we kind of nee $2k $~ $\frac ba$...

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

$x^2 + 2\frac b{2a} + (\frac b{2a})^2 - (\frac b{2a})^2 + \frac ca = 0$

$(x + \frac b{2a})^2 - (\frac b{2a})^2 + \frac ca = 0$

And from there it all comes out in the wash.

Example $x^2 + 16x + 15 = 0$

$x^2 + 2*8x + 15 =0$

$x^2 + 2*8x + 64 - 64 + 15 = 0$

$(x^2 + 2*8x + 64) -64 + 15 = 0$

$(x +8)^2 - 64 +15=0$

$(x+8)^2 - 49 = 0$

$(x+8)^2 = 49$

$(x + 8) = \pm 7$

$x = -8 \pm 7$.

===== Old answer =====

One thing, you have a typo on line 2:

But

$ax^2 +bx + c = 0$

$x^2 + \frac bax + \frac ca = 0$

$x^2 + \frac bax + googadee_bob - googadee_bob + \frac ca = 0$.

($anything + 0 = anything$ and $googadee_bob - googadee_bob = 0$. We can always do this... Now why we would *want to is another story....)

$x^2 + \frac ba x + \frac {b^2}{4a^2} - \frac{b^2}{4a^2} + \frac ca = 0$

(If we can do it for $googadee_bob$ we can do it for $\frac {b^2}{4a^2}$. But why we would want to....)

$(x^2 + \frac ba x + \frac {b^2}{4a^2}) - \frac{b^2}{4a^2} + \frac ca = 0$

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

(Let $M = \frac b{2a}$.)

$(x^2 + 2M + M^2) - - \frac{b^2}{4a^2} + \frac ca = 0$

(we know $(x + M)^2 = x^2 + 2M + M^2$)

$(x + M)^2 -\frac{b^2}{4a^2} + \frac ca = 0$

(Oh, that's why we wanted to do it)

$(x + M)^2 = \frac{b^2}{4a^2} - \frac ca= \frac {b^2}{4a^2} - \frac {4ac}{4a^2}=\frac {b^2 - 4ac}{4a^2} $

$x + M = \frac{\pm \sqrt{b^2 -4ac}}{2a}$

(Okay lets get $M$ back into $\frac b{2a}$ form)

$x + \frac b{2a} = \frac{\pm \sqrt{b^2 -4ac}}{2a}$

$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$