INVESTIGATION: Derive the quadratic formula

The aim of this investigation is to establish the quadratic formula, allowing us to find solutions to any quadratic equation of the form $\displaystyle ax^2+bx+c=0$. Let’s start by using the process of completing the square to solve the equation $\displaystyle x^{2} + 2 x - 8 = 0$. We could solve this equation through factorisation but let’s use completing the square and colour code the terms to follow the process. We can then try to follow the same process to solve $\displaystyle ax^2+bx+c=0$.

This investigation is set out as a fill in the gaps exercise. You may wish to print the page to follow along with the steps or sketch the steps as you go in an exercise book.

Starting with $\displaystyle x^2+2x-8=0$, move the constant term to the right-hand side:

We can represent this problem visually as the sum of the area of a square with side length $\displaystyle x$ and the area rectangle with side lengths $\displaystyle 2$ and $\displaystyle x$, is equal to a square with total area $\displaystyle 8$:

Fill in the missing side lengths.

 

Splitting the green rectangle down the middle we can place it on either side of the blue rectangle. Fill in the missing side lengths and areas.

 

We can “complete the square” on the left by adding the missing corner. We do this to both sides to balance the equation. Fill in the missing area.

 

Combining the terms on the right-hand side to form a new square, we now have equivalent squares. Fill in the missing area for the new rectangle on the right.

 

And hence,

$\displaystyle \left(x + 1\right)^{2} = 9$

So we know the side length $\displaystyle x + 1$ must be equal to the side length of the grey rectangle which is $\displaystyle 3$, so a solution is $\displaystyle x=$___.

If we are looking to solve the equation generally and not just in a concrete case we can also have $\displaystyle x + 1 = - 3$, and hence our second solution is ___.

 

Let’s now repeat this process to deduce the quadratic formula.

Starting with $\displaystyle ax^2+bx+c=0$, divide through by $\displaystyle a$ and move the constant term to the right-hand side:

Let’s again represent this problem visually as the sum of the area of a square with side length $\displaystyle x$ and the area of a rectangle with side lengths $\displaystyle \frac{b}{a}$  and $\displaystyle x$, is equal to a square with total area $\displaystyle -\frac{c}{a}$:

Fill in the missing side lengths and area.

 

Splitting the green rectangle down the middle we can place it on either side of the blue rectangle. Fill in the missing side lengths and areas.

 

We can “complete the square” on the left by adding the missing corner. We do this to both sides to balance the equation. Fill in the missing areas.

 

Combining the terms on the right-hand side to form a new square, we now have equivalent squares.

Fill in the gaps to find a simplified expression for the area of our new grey rectangle:

$\displaystyle -\frac{c}{a}+\frac{b^2}{4a^2}$	$\displaystyle =$	______________________	(make both fractions have the same denominator)
	$\displaystyle =$	______________________	(combine to form a single fraction)

Does the numerator look familiar? Fill in the gap for the area of our new grey rectangle.

 

And hence,

$\displaystyle (x+$____$\displaystyle )^{2}$$\displaystyle =$_________

From here let’s rearrange algebraically:

1. Take the square roots of both sides, don’t forget to indicate there are two answers
2. Simplify the fraction by recalling $\displaystyle \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$
3. Isolate the $\displaystyle x$ on the left by moving the other term to the right
4. Write the right hand side as a single fraction

 

If you have followed the steps correctly you have established quadratic formula:

For a quadratic equation of the form $\displaystyle ax^2+bx+c=0$, solutions are:

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

Further questions to ponder:

When and how was the formula originally established?

Are there other ways to establish the formula?

Are there formulas for cubics, quartics or higher degree polynomials? What is the highest degree polynomial that can be solved in general using a formula?