When $$a$$ equals 1

math
Published

August 20, 2022

## 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!

My reasoning
• $$8 \times -7 = 56$$
• $$8 -7 = 1$$.

## 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.