## The Equation

This is called the standard form of a quadratic equation.

\[ ax^2 + bx + c = 0 \]

You can solve this equation with the quadratic formula:

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

Or by turning it into something that looks like this:

\[ (x + \dots)(x + \dots) = 0 \]

But how exactly do you turn this:

\[ ax^2 + bx + c = 0 \]

into this?:

\[(x + ...)(x + ...) = 0\]

## Noticing a pattern

The first things I noticed were:

- The equation \(ax^2 + bx + c = 0\) was easier to solve in this form: \((x + ...)(x + ...) = 0\)
- Every problem that I was solving was following a pattern: find \(x + b\) and \(xc\)

Every problem was following this pattern, and eventually turned the quadratic equation into a more manageable equation that looked like this: \((x + ...)(x + ...) = 0\). On question 2 or 3 I started to actually notice the pattern, and now, I had a question.

### The Pattern

The pattern was \(y \times z = c\) and \(y + z = bx\). The textbook applied it to almost every question for the quadratic equation

### The Question

My question was if there was a pattern, could I apply it to *all* the problems? Would it work if there was no \(c\)? Would it work if there was no \(bx\) I decided to write my question down in a notebook, and test the theory that I had. Would \(y \times z = c\) and \(y + z = bx\) hold up for all the problems?

## Testing a theory

For the first couple of questions, it worked! Then I came to a very different problem:

\[ 56 - x^2 - x = 0 \]

There was no \(b\)! It was all mixed around and not in the proper form! Would my theory hold up against this?

Yes!

\[ \begin{aligned} 56 - x^2 - x &= 0\\ x^2 + x - 56 &= 0\\ \end{aligned} \]

I was able to put it into the correct form, but how would I account for the missing *b*? This is essentially what would make or break my theory! Then I remembered that if no coefficient is specified the number is 1! I could now solve the equation! My answer was \(x = -7\) and \(x = 8\). It had worked!

## More Testing

Now, it worked for questions with both a \(bx\) and \(c\), either a \(bx\) or \(c\), and questions with neither. I started to test with made-up questions, not in the textbook, because maybe the questions in the textbook were made to work. The first one was:

\[ x^2 + 7x + 12 = 0 \]

That worked.

The next problem was:

\[ x^2 + 11x + 28 = 0 \]

That worked too!

I could not seem to find a problem that disproved my theory!

## Using the theory

Here is the easiest way to solve quadratic equations, without using the quadratic formula.

Here is a problem:

\[ x^2 + 9x + 8 = 0 \]

Now, it looks like there is nothing in common between \(9x\) and \(8\). The numbers that add up to nine are \(1 + 8\), \(2 + 7\), \(3 + 6\), and \(5 + 4\). The factors of eight are \(2 \times 4\) and \(1 \times 8\). The numbers that 9 and 8 have in common are 1 and 8. Those are our two numbers that we need for our more simplified equation. Our new simplified equation is:

\[ (x + 9)(x + 8) = 0 \]

This is much easier to solve than the previous one! Now all we need to do is add the opposite of a positive number, which is a negative number. Our final answer is as such:

\[
\begin{aligned}
(x + 9)(x + 8) &= 0\\
x =& -9\\
x = &-8
\end{aligned}
\] ## Why does \(a\) *have* to equal one?

If you do not recall, the equation is \(ax^2 + bx + c = 0\). There is an \(a\), \(b\), and \(c\). The values for \(b\) and \(c\) are decided by common numbers between the sum and the product of \(y + z\) and \(y \times z\). But this theory only works when \(a = 1\). What happens when \(a = 5\)? If that happens, \(a\) can be split up in many different ways! The theory only works when \(a\) cannot be split up. It is still quite a useful way to solve quadratic equations when \(a\) is equal to one without using the quadratic formula.