Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Activity

1.

Here is one way to make sense of how the quadratic formula came about. Study the derivation until you can explain what happened in each step. Write down your explanation for each step.

Original equation

$a x^{2} + b x + c = 0$

Step 1 - $4 a^{2} x^{2} + 4 a b x + 4 a c = 0$

Step 2 - $4 a^{2} x^{2} + 4 a b x = - 4 a c$

Step 3 - $(2 a x)^{2} + 2 b (2 a x) = - 4 a c$

Step 4 - $M^{2} + 2 b M = - 4 a c$

Step 5 - $M^{2} + 2 b M + b^{2} = - 4 a c + b^{2}$

Step 6 - $(M + b)^{2} = b^{2} - 4 a c$

Step 7 - $\sqrt{(M + b)^{2}} = \sqrt{b^{2} - 4 a c}$

$M + b = \pm \sqrt{b^{2} - 4 a c}$

Step 8 - $M = - b \pm \sqrt{b^{2} - 4 a c}$

Step 9 - $2 a x = - b \pm \sqrt{b^{2} - 4 a c}$

Step 10 -

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

Compare your answer:

Original equation

$a x^{2} + b x + c = 0$

Step 1 - Multiply by $4 a$ so the coefficient of $x^{2}$ is a perfect square.

$4 a^{2} x^{2} + 4 a b x + 4 a c = 0$

Step 2 - Isolate the constant term to the right.

$4 a^{2} x^{2} + 4 a b x = - 4 a c$

Step 3 - Write the squared term as (something) $^{2}$ . Write the linear term as $2 b \cdot$ (something).

$(2 a x)^{2} + 2 b (2 a x) = - 4 a c$

Step 4 - Use $M$ as the placeholder for $2 a x$ .

$M^{2} + 2 b M = - 4 a c$

Step 5 - Add $b^{2}$ to complete the square.

$M^{2} + 2 b M + b^{2} = - 4 a c + b^{2}$

Step 6 - Write the left side as a squared factor and commute the terms on the right.

$(M + b)^{2} = b^{2} - 4 a c$

Step 7 - Find the square roots of each side of the equation.

$\sqrt{(M + b)^{2}} = \sqrt{b^{2} - 4 a c}$

$M + b = \pm \sqrt{b^{2} - 4 a c}$

Step 8 - Subtract $b$ to isolate $M$ .

$M = - b \pm \sqrt{b^{2} - 4 a c}$

Step 9 - Replace $M$ with $2 a x$ .

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

Step 10 - Divide both sides by $2 a$ to isolate $x$ .

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

Are you ready for more?

Extending Your Thinking

1.

Here is another way to derive the quadratic formula by completing the square.

First, divide each side of the equation $a x^{2} + b x + c = 0$ by $a$ to get $x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$ .
Then, complete the square for $x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$ .

In problems 1 – 10, briefly explain what happens in each step by answering the question.

Original equation

$x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$

Step 1
What constant should be subtracted from both sides?

Enter the constant.

Compare your answer:

$- \frac{c}{a}$

Now the equation looks like

$x^{2} + \frac{b}{a} x = - \frac{c}{a}$

2.

Step 2
What constant do we need to add to both sides to complete the square?

Enter the constant.

Compare your answer:

$(\frac{b}{2 a})^{2}$

To complete the square, we add it to both sides. Now the equation looks like

$x^{2} + (\frac{b}{2 a}) x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}$

Notice that the denominators are different on the left side of the equation (for the second and third terms). In order to form a common denominator of $2 a$ , we must multiply the middle term by a factor of $\frac{2}{2}$ .

$x^{2} + \frac{2}{2} (\frac{b}{a}) x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}$

Regrouping the factors in the middle term differently yields

$x^{2} + 2 (\frac{b}{2 a}) x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}$

3.

Step 3
Now we can complete the square. What is the binomial that is squared on the left side?

Enter the binomial.

Compare your answer:

$x + \frac{b}{2 a}$

Now the equation looks like

$(x + \frac{b}{2 a})^{2} = - \frac{c}{a} + \frac{b^{2}}{4 a^{2}}$ .

4.

Step 4
Next, we add the fractions on the right side of the equation. What will the common denominator be?

Enter the common denominator.

Compare your answer:

The common denominator will be $4 a^{2}$ .

Now the equation looks like

$(x + \frac{b}{2 a})^{2} = - \frac{4 a c}{4 a^{2}} + \frac{b^{2}}{4 a^{2}}$ .

5.

Step 5
Once the fractions on the right side of the equation are added together, what will the numerator of that new fraction be?

Enter the numerator.

Compare your answer:

$b^{2} - 4 a c$

Now the equation looks like $(x + \frac{b}{2 a})^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$ .

6.

Step 6
What is the next step necessary to isolate $x$ on the left side?

Enter the next step.

Compare your answer:

Take the square root of both sides.

Remember to put the $\pm$ in front of the radical! Now the equation looks like

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

7.

Step 7
How can the square root be simplified?

Compare your answer:

The square root of a fraction is the square root of the numerator divided by the square root of the denominator.

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

8.

Step 8
What is the denominator of the fraction on the right side after simplification?

Enter the denominator.

Compare your answer:

$2 a$

Now the equation looks like $x + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ .

9.

Step 9
What is the next step I need to take to isolate $x$ on the left side of the equation?

Enter the next step.

Compare your answer:

Subtract $\frac{b}{2 a}$ from both sides.

Now the equation looks like

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

10.

Step 10
What do you notice about the denominator of both fractions on the right side?

Enter your observation.

Compare your answer:

They are the same.

Add the fractions, and we come to the result: the quadratic equation,

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

Video: Deriving the Quadratic Formula

Watch the following video to learn more about deriving the quadratic formula.

Access multimedia content

Self Check

Looking at the steps of deriving the quadratic formula, which step should follow the one below?

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

Subtract $\frac{b}{2a}$ from both sides.
Multiply both sides by $4a^2$ .
Take the square root of both sides.
Complete the square.

Additional Resources

A General Way to Derive the Quadratic Formula

Here is another way to look at how to derive the quadratic formula:

We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 and obtain a positive $a$ . Given $a x^{2} + b x + c = 0$ , $a \neq 0$ , we will complete the square as follows:

Step 1 - First, move the constant term to the right side of the equal sign:

$a x^{2} + b x = - c$

Step 2 - As we want the leading coefficient to equal 1, divide through by $a$ :

$x^{2} + \frac{b}{a} x = - \frac{c}{a}$

Step 3 - Then, find $\frac{1}{2}$ of the middle term, and add $(\frac{1}{2} \cdot \frac{b}{a})^{2} = \frac{b^{2}}{4 a^{2}}$ to both sides of the equal sign:

$x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} = \frac{b^{2}}{4 a^{2}} - \frac{c}{a}$

Step 4 - Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

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

Step 5 - Now, use the square root property, which gives:

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

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

Step 6 - Finally, add $- \frac{b}{2 a}$ to both sides of the equation and combine the terms on the right side. Thus:

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

Try it

Try It: A General Way to Derive the Quadratic Formula

Using the general way to derive the quadratic formula, find the step after

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

Here is how to find the next step:

When deriving the quadratic formula, the object is to get the $x$ alone.

After $x + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}}$ , the $\frac{b}{2 a}$ must be subtracted from each side. Then $x = - \frac{b}{2 a} \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}}$ can be simplified.

9.8.2 Deriving the Quadratic Formula, Part 2

Activity

Are you ready for more?

Extending Your Thinking

Video: Deriving the Quadratic Formula

Additional Resources

A General Way to Derive the Quadratic Formula

Try It: A General Way to Derive the Quadratic Formula