Remember the quadratic formula? That preposterous equation you were taught in middle school or high school that didn’t even seem like an equation anymore. If you don’t remember what I’m talking about, then here look at it in its full glory!

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

To be honest, while this formula seems perfectly good when we are directly looking at coefficients like those below, I want to propose a more simpler and elegant formula that has more relevance and meaning to an actual quadratic (because that equation is just too long).

** Introducing**

x = h \pm \sqrt{-\frac{k}{a}}

## Wait what? Why?

Now, you might be confused by what h and k are, but they are simply coordinates of the vertex of the parabola (h,k).

What this equation shows us is that all the information about a parabola is contained in the placement of its vertex and its a value. This should make sense as any two parabolas can have the same vertex, but not have the same roots. It is the a value that dictates the roots they have.

Now what makes this form of the quadratic formula so elegant?

Part of the reason is that we can compute it much more quickly because all we have to do is a simple conversion of a normal quadratic into vertex form, namely y = a(x-h)^2 + k, and then from there, all you have to do is pretty much read off a,k, and h from the formula and put it in the quadratic formula.

Also, by converting your quadratic into vertex form, you also get more crucial information about your quadratic by acquiring the vertex on top of the roots. As long as you can convert a quadratic to vertex form quickly, this formula gives more information about the quadratic and is also much simpler. Now, how do we compute the vertex form of a quadratic quickly?

**Computing Vertex Form of A Quadratic Quickly**

Consider the quadratic y = x^2 + 2x + 7.

**I already know the vertex form of this, but how?!**

I have a trick. Whenever you have a quadratic like this, what you need to do is take the x term (in this case 2x) and divide its coefficient by 2 and rewrite your current formula as such (because 2/2 = 1)

y = (x+1)^2

Next, all you have to do is square that 1 value, so that is 1 and subtract your squared value (1) from the last term, namely 7, giving you 6. All you have to do is simply tack it on.

y = (x+1)^2 + 6

**And that’s it! You have your vertex formula and guess what? **

**You also have your roots, namely, by plugging into x = h \pm \sqrt{-\frac{k}{a}}**,

x = -1 \pm \sqrt{-6}

There is one small caveat, if you have a formula like 2x^2 + 4x + 9, you have to factor out the a value (2), and rewrite it as 2(x^2+2x+4.5) before continuing on.

As you can see converting into vertex form is not as hard as it seems, and you get more information by doing the quadratic formula this way!

**Summing Up**

Now while I do personally like my quadratic formula and think it is useful when looking at a quadratic in a certain way, by no means, if you are learning the quadratic formula in its traditional form just ignore it. All teachers will expect it from you, so know it! I’m just showing how we can’t take things for granted either when there could be simpler forms or intuition in another form of something. It’s more about the idea of being creative that’s important than simply thinking math is a long list of rules! It’s not! We do math for fun and to explore, not to regurgitate information in non-intuitive ways. Stay safe and stay creative!

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**Resources**

https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html

https://www.mathsisfun.com/algebra/quadratic-equation-graphing.html