Lesson 32 – Solving Quadratic Equations By Factoring

What is a Quadratic Expression? One where the variable is squared. It can take two forms. Here is the first: 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄To turn it into a Quadratic Equation, add a “= 0” at the end, so it looks like this: 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎 This is called Standard Form

There are four algebraic methods that can be used to solve a quadratic equation, meaning, find all values for 𝒙 :1) By factoring 2) By using the quadratic formula3) By using the Square Root Property4) By completing the square. This one we won’t be covering.

In order to teach the first one, we must first explain the Zero Product Property:If two terms multiplied together equal 0, then at least one of them had to be zero in the first place. Maybe both!

So…

If 𝒙𝒚 = 𝟎, then either 𝒙 or 𝒚 is 0.If 𝟔𝒙 = 𝟎, then 𝒙 must be 0.If 𝒙(𝒙 + 𝟐) = 𝟎, then either 𝒙 is 0, or (𝒙 + 𝟐) is 0. (In order for 𝒙 + 𝟐 to be 0, what does 𝒙 have to be?)If 𝒙 − 𝟏 (𝒙 + 𝟐) = 𝟎, then either (𝒙 − 𝟏) or (𝒙 + 𝟐) is 0.

In order for 𝒙 − 𝟏 to be 0, what does 𝒙 have to be? _____In order for 𝒙 + 𝟐 to be 0, what does 𝒙 have to be? _____ >>> CLASSWORK <<<

Let’s combine the concepts. What if you’re given the following Quadratic Equation (in Standard form) –𝒙𝟐 + 𝒙 − 𝟐 = 𝟎and told to solve it, meaning, find all possible values of 𝒙 ?

To solve by factoring:

Step 1: Factor it: If 𝒙𝟐 + 𝒙 − 𝟐 = 𝟎Then (𝒙 + 𝟐)(𝒙 − 𝟏) = 𝟎 This is the 2nd form of a Quadratic Equation, called Factored Form

Step 2: Use the Zero Product Property to determine all values for 𝒙

If (𝒙 + 𝟐)(𝒙 − 𝟏) = 𝟎, then either (𝒙 + 𝟐) = 𝟎 or (𝒙 − 𝟏) = 𝟎 (or both)One at a time: If (𝒙 + 𝟐) = 𝟎 then solving for x, gives you 𝒙 = −𝟐

If (𝒙 − 𝟏) = 𝟎 then solving for x, gives you 𝒙 = 𝟏

So…There are two solutions determined by factoring: −𝟐 and 𝟏These values make the equation true.How do you know if they’re correct? Substitute them back in and check.If 𝒙𝟐 + 𝒙 − 𝟐 = 𝟎 ………. (−𝟐)𝟐+ −𝟐 − 𝟐 = 𝟒 − 𝟐 − 𝟐 = 𝟎 YES!If 𝒙𝟐 + 𝒙 − 𝟐 = 𝟎 ………. (𝟏)𝟐+ 𝟏 − 𝟐 = 𝟏 + 𝟏 − 𝟐 = 𝟎 YES!

BTW… They’re also called the ROOTS of the equation.

A quadratic equation can have 0 solutions/roots, 1 solution/root, or 2 solutions/roots.

Here’s an example where it has 0: 𝒙𝟐 + 𝟐 = 𝟎To solve for 𝒙, subtract 2 from both sides, to get 𝒙𝟐 = −𝟐How is it possible to square a number, and the result is negative? It ISN’T possible, so there are no solutions/roots, meaning no value for 𝒙 that makes it true.

Here’s an example where it has 1: 𝒙𝟐 + 𝟐𝒙 + 𝟏 = 𝟎If we factor it, we get 𝒙 + 𝟏 𝒙 + 𝟏 = 𝟎Using the Zero Product Property, either 𝒙 + 𝟏 = 0 or 𝒙 + 𝟏 = 𝟎But wait! They’re both the same!So the solution is only –1.

Here’s an example where it has 2: 𝒙𝟐 + 𝟏𝟏𝒙 + 𝟑𝟎 = 𝟎If we factor it, we get 𝒙 + 𝟓 𝒙 + 𝟔 = 𝟎Using the Zero Product Property, either 𝒙 + 𝟓 = 0 or 𝒙 + 𝟔 = 𝟎So the solutions are –5 and –6.

𝒙𝟐 + 𝟏𝟏𝒙 + 𝟑𝟎 = 𝟎 30 𝑥 + 6 (𝑥 + 5) –6 , –5

𝒙𝟐 + 𝟏𝟐𝒙 + 𝟐𝟎 = 𝟎

𝒏𝟐 + 𝟐𝒏 − 𝟖 = 𝟎

𝒙𝟐 + 𝒙 − 𝟏𝟐 = 𝟎

𝟐𝒙𝟐 − 𝟐𝟐𝒙 + 𝟒𝟖 = 𝟎

𝟑𝒌𝟐 + 𝟐𝟏𝒌 + 𝟑𝟔 = 𝟎

𝒙𝟐 − 𝟗 = 𝟎

𝟖𝟏𝒙𝟐 − 𝟒𝟗 = 𝟎

𝟗𝒙𝟐 − 𝟏 = 𝟎

𝟖𝟏𝒙𝟒 − 𝟏𝟔 = 𝟎

Given these expression, solve each for 𝒙 by factoring. Hint: We factored them in Lessons 30 and 31; just look them up…

Quadratic Equation Lesson Factors Solution(s)/Root(s) -

>>> CLASSWORK <<<

𝟒𝒙𝟐 + 𝟏𝟎𝒙 + 𝟔 = 𝟎

𝟔𝒙𝟐 + 𝟏𝟗𝒙 + 𝟏𝟎 = 𝟎

𝟔𝒙𝟐 − 𝟐𝟏𝒙 − 𝟒𝟓 = 𝟎

Given these expression, solve each for 𝒙 by factoring. Hint: We factored them in Lessons 30 and 31; just look them up…

Quadratic Equation Lesson Factors Solution(s)/Root(s) -

Solve by factoring.

What is a Quadratic Function? Just put function notation in front – for example - 𝒇(𝒙).𝒇 𝒙 = (𝒙 + 𝟐)(𝒙 − 𝟏) This is the factored form

The equivalent function is 𝒈 𝒙 = 𝒙𝟐 + 𝒙 − 𝟐 This is the Standard form (it’s those binomials, multiplied out)The solutions to both are (−𝟐 and 𝟏). They are also called the “ZEROES”, since they cause the function to become 0.To find the zeroes of a function in standard form, factor it.

What are the zeroes of these functions?

𝒇 𝒙 = 𝒙 − 𝟏𝟓 (𝒙 + 𝟒) 𝒇 𝒙 = −𝟒(𝒙 − 𝟖) 𝒇 𝒙 = 𝟐𝒙𝟐 + 𝟏𝟓𝒙 + 𝟐𝟓

Solve by factoring.

𝟓𝒙𝟐 = 𝟒𝟓 𝒙 + 𝟐 𝒙 + 𝟖 = −𝟓𝟒𝒙𝟐 − 𝟑𝟔𝒙 = 𝟎

>>> CLASSWORK <<<

>>> CLASSWORK <<<

P. 963

Here are two squares and their areas. In terms of x, how much taller is the left square than the right one? Explain your reasoning.

𝒙𝟐 + 𝟏𝟐𝒙 + 𝟑𝟔16

𝒉 = −𝟏𝟔𝒕𝟐 + 𝒗𝒕 + 𝒔

The height 𝒉 of an object traveling through the air at any moment is given by this equation:

Where 𝒕 is the amount of time that has passed since the object began its journey𝒗 is the initial velocity𝒔 is the initial height