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A 2.7 Solving Quadratic Equations by Graphing and by Factoring

Chapter 2 Section A7

Algebra Topics Solving Quadratic Equations by Guess and Check, Graphing, or Factoring

Solving Quadratic Equations by Guess and Check and by Graphing

Quadratic equations are equations that have x^2 as the highest power of x. For example, x^2 = 25 is a quadratic equation, as is  x^2-6x+8=3. To solve these equations, let’s first try the strategy of guess and check to see what will work to make the equation true.

Example 1 For the equation x^2 = 25, what numbers will work?

We can see that letting x = 5 will make the equation true, because 5^2 = 25. But another number will also make this equation true: x = -5, because (-5)^2= 25. This means that the equation has two solutions.

Example 2 For the equation x^2-6x+8=3, what numbers will work?

For the equation x^2-6x+8=3, the guessing is more difficult. We might first try guessing an easy number like zero:
(0)^2 - 6(0) + 8 = 0 - 0 + 8 = 8, which is not correct, since we are trying to get 3. We can next try guessing higher or lower than 0. Since 1 is the next easiest number to try, let’s try that.

(1)^2 - 6(1) + 8 = 1 - 6 + 8 = 3, which is correct!

Since the other equation had two solutions, we can try to find a second solution for this equation as well. Some searching will reveal that the second number that works is x = 5.

Our guess and check is similar to making a table of points for the equation  y = x^2-6x+8, and looking for when we get y = 3:

image

We see that there are two places (circled) where y = 3.

Example 3 For the equation x^2-6x+8=3 and x^2 = 25, where can we find the solutions on each graph?

We can graph the equation y=x^2-6x+8 using the table of points in the previous example to see where the graph passes through y = 3. Notice that a quadratic equation graphs a parabola shape, not a line, because squaring creates two possible solutions, as we saw from our guess and check work. The graph below shows the solution to y=x^2-6x+8 = 3 happens at (1,3) and (5,3). Thus the solutions are x = 1 and x = 5.

Similarly, a graph of y = x^2 shows two points where y = 25, at x = 5 and x = -5:

Thus, the solutions are x = 5 and x = -5.

Solving Quadratic Equations Using the Zero Factor Principle

Another way to solve a  equation is to use the zero factor principle. Let’s look first at what the zero factor principle tells us.

If ab = 0, then what do you know about what a must equal? What about b?

You know that one of the two variables must be equal to zero. For example, we could have a = 0 and b = 5, because 0*5 = 0, or we could have a = 2 and b = 0, because 2*0 = 0. Or both a and b could equal zero, since 0*0 = 0

Thus, if ab = 0, then a must be zero, or b must be zero, or both.

If two things multiply to equal zero, one or both of them must be zero.
Example 4 To use the zero factor prinicple with the equation x^2-6x+8=3, 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.

Next, to solve, we need to factor the left side so that we have two factors (two things that are multiplied) that equal zero.

Reminder: to factor, set up (x    +   ?)(x  +    ?) because (x )(x )=x^2
Next, 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

This means that either the first factor is equal to zero, or the second factor is equal to zero, or both, so x - 1 = 0 or x - 5 = 0, or both.

Solve each one:

x - 1 = 0

+1    +1

x = 1

x - 5 = 0

+5   +5

x = 5

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

To use this same idea with the equation x^2=25, we subtract 25 from both sides so that the equation is equal to zero:

We get x^2 - 25 = 0.
Next, to solve, we need to factor the left side so that we have two factors (two things that are multiplied) that equal zero.

Reminder: to factor, set up (x    +   ?)(x  +    ?) because (x )(x )=x^2

Note: since there is no middle term, it might help to write the equation as x^2 + 0x - 25 = 0, to show that the middle term is zero. So we are looking for two numbers that multiply to get -25 and add to get 0. The numbers 5 and -5 multiply to equal -25 and add to get 0.

x^2 + 0x - 25 = 0 factors to

(x - 5)(x + 5) = 0

Using the zero factor principle, this means we know that x - 5 = 0 or x + 5 = 0. Solve each:

x - 5 = 0

+5   +5

x = 5

and

x + 5 = 0

-5   -5

x = -5

These are the same x values we found in our guess and check and on the graph.

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College Mathematics for Elementary Education with Algebra Extensions Copyright © by Kathleen Offenholley and Fatima Prioleau. All Rights Reserved.

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