Every student of mathematics will eventually come across the quadratic formula in some algebra class. A student will probably see something along the lines of:

The solutions to the equation $ax^2+bx+c=0,$ where $a \neq 0,$ are $x = \dfrac{-b \pm \sqrt{{b^2-4ac}}}{2a}.$

But where does this equation even come from? Most students don’t see a proof in their algebra classes (and may never come across one for the rest of their math careers), but in this post, we’ll go through exactly that.

Consider the quadratic $ax^2+bx+c,$ where $a \neq 0.$ We can assume that $a \neq 0,$ since if $a = 0,$ then $ax^2+bx+c = bx+c,$ which is no longer a quadratic (it is now linear). It is very important to state that $a \neq 0,$ since the derivation of the quadratic formula will use divisions by $a$ (recall that you can divide by a number as long as it is not 0). An important concept that the proof requires is completing the square, which is covered in this post. Now, here we go!

We first start with our initial problem:

$$ax^2+bx+c=0$$

Next, we’ll factor out an $a$ from the first two terms (The $ax^2$ and $bx$ terms). The reason we do this is to set ourselves up for completing the square. When we complete the square, we want to work with an $x^2$ term with a coefficient of $1.$ It may seem a bit odd to factor out an $a$ from a term that doesn’t even have an $a,$ but it’s actually still possible! It’s very similar to rewriting $3$ as $2 \cdot \frac{3}{2}.$

$$a\left(x^2+ \dfrac{b}{a}x \right) + c = 0$$

If you’re unsure, you can always distribute back the $a$ and check that it indeed matches the equation that we started with!

Now, we’ll be completing the square. Remember that when completing the square, you take the coefficient of the $x$ term, half it, and then square it. In this case, the coefficient of the $x$ term is $\frac{b}{a}.$ If we take half of this, we get $\frac{b}{2a}.$ Finally, if we square this term, we get $\frac{b^2}{4a^2}.$ We’re now going to add this to the inside of the parenthesis.

However, remember that you’re not allowed to just add a value to one side of the equation! Doing so will cause the equation to become invalid. If you add a value to one side of an equation, you must at the same value to the other side. Be careful, though! We’re not just adding $\frac{b^2}{4a^2} ;$ we’re actually adding $a \cdot \frac{b^2}{4a^2} = \frac{b^2}{4a}$ because of the $a$ that we factored out in the previous step. We now have the following:

$$a\left(x^2+ \dfrac{b}{a}x + \dfrac{b^2}{4a^2} \right) + c = \dfrac{b^2}{4a}$$

Since we just completed the square, we can factor! We’re also going to be subtracting $c$ from both sides.

$$a \left(x+\dfrac{b}{2a} \right)^2 = \dfrac{b^2}{4a} - c$$

Again, if you’re unsure, you can distribute $\left(x+\frac{b}{2a} \right)^2$ and check that it matches the corresponding portion from the previous step.

Next, we’re going to express the right side as a single fraction. We can do so by multiplying $c$ by $\frac{4a}{4a}=1.$ Remember that multiplying by $1$ doesn’t change the value of something!

$$a \left(x+\dfrac{b}{2a} \right)^2 = \dfrac{b^2-4ac}{4a}$$

Now, we’ll divide by $a$ on both sides. Recall that we have no issues with $a$ as a denominator since we know that $a \neq 0.$

$$\left(x+\dfrac{b}{2a} \right)^2 = \dfrac{b^2-4ac}{4a^2}$$

Next, we’ll take the square root of both sides. Remember that when taking the square root, you must consider plus or minus.

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

We can simplify further by using the following property for square roots: $\sqrt{\frac{\alpha}{\beta}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}}.$

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

Now, we’ll simplify $\sqrt{4a^2}.$ We’ll be using the property that $\sqrt{\alpha\beta} = \sqrt{\alpha}\sqrt{\beta}.$

$\sqrt{4a^2} = \sqrt{4}\sqrt{a^2}=2\sqrt{a^2}.$ You might be tempted to say that $\sqrt{a^2}=a,$ but this is actually not the case! It turns out that $\sqrt{a^2} = |a|.$ For instance, if we consider $a=-2,$ then $\sqrt{a^2} = \sqrt{(-2)^2} = \sqrt{4} = 2.$

If we now come back to where we left off, we have the following:

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

Because of the $\pm$ , we can actually get rid of the absolute value around the $a.$ This is because $\pm|a| = \{a,-a\} = \pm a.$ Don’t worry too much if this step is a bit confusing. All we need to know is that we can get rid of the absolute value.

We’re now left with the following:

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

We’re almost there! The next step is to subtract $\frac{b}{2a}$ from both sides.

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

Since both fractions have the same denominator, we can easily combine them, and we’ll have the quadratic formula!

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

The proof is now complete!