## Derivation of the Quadratic Formula

When we discuss about functions, we usually talk about their roots, or geometrically where their graphs pass through the x-axis. For example, $- 3, 1$ and $3$ are the roots of the graph of the function in Figure 1 because it passes through $x = -3, x =1$ and $x=3$.

Since we are looking for points on the x-axis, it means that all the points that we are looking for have $y$-coordinate $0$. As a consequence, (i) if we want to find the root of a quadratic function we have set $y = 0$ and then solve for the values of $x$.

Figure 1 – The x-axis and the line with equation y = 0 are basically the same line so all points on the x-axis have y-coordinate 0.

With the things above in mind, let us find the roots of two quadratic functions: $(1) y = x^2 + 5x + 6$ and $(2) y = x^2 + 3x + 1$.

Finding the root by Factoring

The roots of the function $y = x^2 + 5x + 6$ are easy to find. As we have said, to get the root of a function, we set $y$ to $0$ and then find the value the value of $x$. Solving by factoring, we have

$x^2 + 5x + 6 = 0 \Rightarrow (x + 2)(x + 3) = 0$

Now, $(x + 2)(x + 3) = 0$ if $x + 2 = 0$ or/and $x + 3 = 0$. Solving for the value of $x$ on both equations, we have $x= -2$ and $x = -3$.

Thus, even though we have not seen the graph of the function yet $y = x^2 + 5x + 6$, we are sure that it will pass through $x = -2$ and $x = -3$. If you want to verify if the graph of the function $y= x^2 + 5x + 6$ indeed passes through the $x$-axis at $x = -2$ and $x = -3$, you can verify its graph using a graphing software or a graphing calculator.

Figure 2 – We are sure that the graph of $y = x^2 + 5x + 6$ will pass through the yellow points.

Factoring: Another Interpretation

Another possible interpretation of the expression $x^2 + 5x + 6$ can be the area of a rectangle with length $x + 3$ and width $x+2$. The distribution of the products of the terms of the expressions are represented by the four rectangles formed shown below.

Figure 3 – Another interpretation of the expression x^2 + 5x + 6.

Let us try another example: Let us find the roots of the quadratic function $y = x^2 + 3x + 1$. You will probably observe that there is no way that we can factor this expression. The last term is $1$, but there are only two factors of 1: $\{1,1\}$ and $\{-1,-1\}$, so this means that that the numerical coefficient of$x$ must be $2$ or $-2$, but it is equal to $3$. Hence, the expression is not factorable.

Since, the expression is not factorable, we cannot find the length and width of a rectangle with area $x^2 + 3x + 1$. The easiest way probably to find its length and width is to assume that it is a square.

Completing the Square

We have a quadratic expression which we assumed a perfect square so its factor must be of the form $(x+b)(x+b)$ where $b$ is a real number. Also, $(x+b)^2 = x^2 + 2bx + b^2$. If we consider $x + b$ as a side of the square, then the product of the expressions will form two squares namely $x^2$ and $b^2$, and 2 congruent rectangles with each having an area of $bx$.

Figure 4 – A square with side x + b, where b is a constant.

If we want to use the product of $x+ b$ above, first, we have to take off $x^2$ as one of the squares. Then we are left with a figure with area $3x$ which we will divide into two congruent rectangles. If we are going to follow the positions of the rectangles in Figure 4, then we will have an $x^2$ and two pieces of $\frac{3}{2}x$ (see Figure 5).To construct a square, we extend one of the sides of each of the congruent rectangles.

Figure 5: Completing the square of x^2 + 3x + 1.

Since we have two small rectangles with area $\frac{3}{2}x$, and the longer side (in the diagram) is $x$, it follows that the other dimension is $\frac{3}{2}$ which gives us a smaller square with area $\frac{9}{4}$.

Figure 6 – The area of the small formed is 9/4 and the side length of the big square formed is x + 3/2.

The biggest square formed in Figure 6 has area $x^2 + \frac{3}{2}x +\frac{9}{4}$, which is $\frac{5}{4}$ more than our original quadratic expression, so we will deduct $\frac{5}{4}$ to preserve the original expression. So our final expression is $x^2 + \frac{3}{2}x +\frac{9}{4} - \frac{5}{4}$

Algebraically, if we have the expression $x^2 + bx + c$, and we want to “compete its square”, we want to transform it to an expression of the form $(x-h)^2 + k$. For example, $x^2 + 6x + 12$ can be expressed as $(x + 3)^2 + 3$. Another example is that $x^2 + 8x + 12$ can be written as $(x + 4)^2 - 4$. Note that the coefficient of $x^2$ should be $1$ so that we are sure that a square $x$ by $x$ is formed as shown in figures 4,5 and 6. In general, the possible steps that we are going to create using the general equation $y = ax^2 + bx + c$ is to set $y$ to $0$ and then find the value of $x$.

In constructing the square in Figure 6, we went through the following processes:

(ii) We made sure that the numerical coefficient of $x^2$ is $1$ to ensure that we have a square with factors (side length) $x$.

(iii) We isolated the constant term $c = 1$, and we just used the first $x^2$ and the second term $3x$.

(iv) To get the area of the smaller square, we divided the numerical coefficient of the $3x$ by $2$ then squared it to get $\frac{9}{4}$.

Shown in Figure 7 is the summary of the steps we did to get the roots of the quadratic function $y = x^2 + 3x + 1$. The rightmost column of the table shows the generalization of our steps, which is getting the roots of the quadratic function $y = ax^2 + bx + c$.

Figure 7 – A step-by-step derivation of the quadratic formula.

The formula $x = \displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ located at the bottom part of the rightmost column of the table in Figure 7 is called the quadratic formula. We have derived the quadratic formula from completing the square of a quadratic equation. From the formula, the roots o the quadratic function $y = ax^2 + bx + c$ are $\displaystyle\frac{-b + \sqrt{b^2-4ac}}{2a}$ and $\displaystyle\frac{-b - \sqrt{b^2-4ac}}{2a}$.