Quadratic Shortcuts

The Equation

This is called the standard form of a quadratic equation.

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

You can solve this equation with the quadratic formula:

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

Or by turning it into something that looks like this:

$(x + \dots) (x + \dots) = 0$

But how exactly do you turn this:

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

into this?:

$(x + . . .) (x + . . .) = 0$

Noticing a pattern

The first things I noticed were:

The equation $a x^{2} + b x + c = 0$ was easier to solve in this form: $(x + . . .) (x + . . .) = 0$
Every problem that I was solving was following a pattern: find $x + b$ and $x c$

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 $y + z = b x$ . 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 $b x$ I decided to write my question down in a notebook, and test the theory that I had. Would $y \times z = c$ and $y + z = b x$ 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!

My reasoning

$8 \times - 7 = 56$
$8 - 7 = 1$ .

More Testing

Now, it worked for questions with both a $b x$ and $c$ , either a $b x$ 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:

$x^{2} + 7 x + 12 = 0$

That worked.

The next problem was:

$x^{2} + 11 x + 28 = 0$

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:

$x^{2} + 9 x + 8 = 0$

Now, it looks like there is nothing in common between $9 x$ 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 $a x^{2} + b x + c = 0$ . 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.