A 2.7 Solving Quadratic Equations by Graphing and by Factoring

Chapter 2 Section A7 NOT YET COMPLETE

Algebra Topics –  Solving Quadratic Equations by Graphing and by Factoring.

Solving Quadratic Equations

Quadratic equations are equations that have x² as the highest power of x. For example, is a quadratic equation, as is  . To solve these equations, let’s first try the strategy of using guess and check to see what will work to make the equation true.

For the equation , we can see that letting will make the equation true, because . But another number will also make this equation true: , because . This means that the equation has two solutions.

For an equation like  , there are several ways to solve for x. One way is to find when the equation $y=x^2-6x+8$ passes through y = 3.

We can see from the graph below that this happens at (1,3) and (5,3). So the solutions are x = 1 and x = 5.

We can also see in the table for $y=x^2-6x+8$ that there are two places where y is 3.

The other way to solve the equation $x^2-6x+8=3$ is to use the zero factor principle, that if ab = 0, then a must be zero, or b must be zero, or both. That is, if two things multiply to equal zero, one or both of them must be zero.

To use this idea, we subtract 3 from both sides so that the equation is equal to zero:

$x^2-6x+8=3$

$-3$      $-3$

We get

$x^2-6x+5=0$

To solve this, we need to factor the left side so that we have two factors that equal zero.

Reminder: to factor, set up (x       ?)(x      ?) because $\left(x\right)\left(x\right)=x^2$

Now find two numbers that multiply to get positive 5 and add to get -6. The numbers -1 and -5 multiply to equal positive 5 and add to get -6.

We get

$x^2-6x+5=0$

$(x-1)(x-5)=0$

So that means that one (or both) of those factors must equal zero:

$x-1=0$ or  $x-5=0$

So x = 1 or x = 5.

These are the same x values we found on the table and in the graph.