Introduction to Qudratic Trinomials and Equations
Quadratic Trinomial
A quadratic trinomial contains an x^{2} term as well as an x term. For example, x^{2} – 6x + 8 is a quadratic trinomial. You can factor quadratic trinomials by using the FOIL method in reverse.
Example
Let's factor x^{2} – 6x + 8.
Start by looking at the last term in the trinomial: 8. Ask yourself, "What two integers, when multiplied together, have a product of positive 8?" Make a mental list of these integers:
Next look at the middle term of the trinomial: –6x. Choose the two factors from the list you just made that also add up to the coefficient –6:
–2 and –4
Now write the factors using –2 and –4:
(x – 2)(x – 4)
Use the FOIL method to double-check your answer:
(x – 2)(x – 4) = x^{2} – 6x + 8
You can see that the answer is correct.
Quadratic Equations
A quadratic equation is an equation that does not graph into a straight line. The graph will be a smooth curve. An equation is a quadratic equation if the highest exponent of the variable is 2. Here are some examples of quadratic equations:
x^{2} + 6x + 10 = 0
6x^{2} + 8x – 22 = 0
A quadratic equation can be written in the form: ax^{2}+ bx + c = 0. The a represents the number in front of the x^{2} variable. The b represents the number in front of the x variable and c is the number. For instance, in the equation 2x^{2} + 3x + 5 = 0, the a is 2, the b is 3, and the c is 5. In the equation 4x^{2} – 6x + 7 = 0, the a is 4, the b is –6, and the c is 7. In the equation 5x^{2} + 7 = 0, the a is 5, the b is 0, and the c is 7. In the equation 8x^{2} – 3x = 0, the a is 8, the b is –3, and the c is 0. Is the equation 2x + 7 = 0 a quadratic equation? No! The equation does not contain a variable with an exponent of 2. Therefore, it is not a quadratic equation.
Solving Quadratic Equations Using Factoring
Why is the equation x^{2} = 4 a quadratic equation? It is a quadratic equation because the variable has an exponent of 2. To solve a quadratic equation, first make one side of the equation zero. Let's work with x^{2} = 4.
Subtract 4 from both sides of the equation to make one side of the equation zero: x^{2} – 4 = 4 – 4. Now, simplify x^{2} – 4 = 0. The next step is to factor x^{2} – 4. It can be factored as the difference of two squares: (x – 2)(x + 2) = 0.
If ab = 0, you know that either a or b or both factors have to be zero because a times b = 0. This is called the zero product property, and it says that if the product of two numbers is zero, then one or both of the numbers have to be zero. You can use this idea to help solve quadratic equations with the factoring method.
Use the zero product property, and set each factor equal to zero: (x – 2) = 0 and (x + 2) = 0.
When you use the zero product property, you get linear equations that you already know how to solve.
Solve the equation: |
x – 2 = 0 |
Add 2 to both sides of the equation. |
x – 2 + 2 = 0 + 2 |
Now, simplify: |
x = 2 |
Solve the equation: |
x + 2 = 0 |
Subtract 2 from both sides of the equation. |
x + 2 – 2 = 0 – 2 |
Simplify: |
x = –2 |
You got two values for x. The two solutions for x are 2 and –2. All quadratic equations have two solutions. The exponent 2 in the equation tells you that the equation is quadratic, and it also tells you that you will have two answers.
Tip: When both your solutions are the same number, this is called a double root. You will get a double root when both factors are the same.
Before you can factor an expression, the expression must be arranged in descending order. An expression is in descending order when you start with the largest exponent and descend to the smallest, as shown in this example: 2x^{2} + 5x + 6 = 0.
All quadratic equations have two solutions. The exponent of 2 in the equation tells you to expect two answers.
Example
x^{2} – 3x – 4 = 0 |
Factor the trinomial x^{2} –3x – 4. |
(x – 4)(x + 1) = 0 |
Set each factor equal to zero. |
x – 4=0 and x + 1= 0 |
Solve the equation. |
x – 4 = 0 |
Add 4 to both sides of the equation. |
x – 4 + 4 = 0 + 4 |
Simplify. |
x = 4 |
Solve the equation. |
x + 1 = 0 |
Subtract 1 from both sides of the equation. |
x + 1 – 1 = 0 – 1 |
Simplify. |
x = –1 |
The two solutions for the quadratic equation are 4 and –1. |
Tip: When you have an equation in factor form, disregard any factor that is a number and contains no variables. For example, in 4(x – 5) (x + 5) = 0, disregard the 4. It will have no effect on your two solutions.
Solving Quadratic Equations by Using the Zero Product Rule
If a quadratic equation is not equal to zero, rewrite it so that you can solve it using the zero product rule.
Example
If you need to solve x^{2} – 11x = 12, subtract 12 from both sides:
x^{2} – 11x – 12 = 12 – 12
x^{2} – 11x – 12 = 0
Now this quadratic equation can be solved using the zero product rule.
A quadratic equation must be factored before using the zero product rule to solve it.
Example
To solve x^{2} + 9x = 0, first factor it:
x(x + 9) = 0
Now you can solve it.
Either x = 0 or x + 9 = 0.
Therefore, possible solutions are x = 0 and x = –9.
Graphs of Quadratic Equations - Parabolas
Introduction to Parabolas
The (x,y) solutions to quadratic equations can be plotted on a graph. These graphs are called parabolas. Typically you will be presented with parabolas given by equations in the form of y = ax^{2} + bx + c.
Notice that the equation y = x^{2} conforms to this formula—both b and c are zero.
y = (1)x^{2} + (0)x + (0) is equivalent to y = x^{2}
The value of a cannot equal zero, however.
Movement of the Parabola on the Graph - Opening Up and Down
If a is greater than zero, the parabola will open upward. If a is less than zero, the parabola will open downward.
The x-coordinate of the turning point, or vertex, of the parabola is given by:
You can use this x-value in the original formula and solve for y (the y-coordinate of the turning point).
There will also be a line of symmetry given by:
For the graph y = x^{2}, x = = 0. The line of symmetry is x = 0. The y-coordinate of the vertex is at located at y = x^{2} = 0^{2} = 0, so the vertex is at (0,0). Technically a parabola could also be given by the formula x = ay^{2} + by + c.
The graph of the equation y = x^{2} is a parabola.
Because the x-value is squared, the positive values of x yield the same y-values as the negative values of x. The graph of y = x^{2} has its vertex at the point (0,0). The vertex of a parabola is the turning point of the parabola. It is either the minimum or maximum y-value of the graph. The graph of y = x^{2} has its minimum at (0,0). There are no y-values less than 0 on the graph.
Movement of the Parabola on the Graph - Moving Up, Down, Left, and Right
The graph of y = x^{2} can be translated around the coordinate plane. While the parabola y = x^{2} has its vertex at (0,0), the parabola y = x^{2} – 1 has its vertex at (0,–1). After the x term is squared, the graph is shifted down one unit. A parabola of the form y = x^{2} – c has its vertex at (0,–c) and a parabola of the form y = x^{2} + c has its vertex at (0,c).
The parabola y = (x + 1)^{2} has its vertex at (–1,0). The x-value is increased before it is squared. The minimum value of the parabola is when y = 0 (because y = (x + 1)^{2} can never have a negative value). The expression (x + 1)^{2} is equal to 0 when x = –1. A parabola of the form y = (x – c)^{2} has its vertex at (c,0) and a parabola of the form y = (x + c)^{2} has its vertex at (0,–c).
What are the coordinates of the vertex of the parabola formed by the equation y = (x – 2)^{2} + 3?
To find the x-value of the vertex, set (x – 2) equal to 0: x – 2 = 0, x = 2. The y-value of the vertex of the parabola is equal to the constant that is added to or subtracted from the x squared term. The y-value of the vertex is 3, making the coordinates of the vertex of the parabola (2,3).
If parabolas with the formula y = x^{2} + bx + c open upward or downward, how do you think parabolas given by the formula x = ay^{2} + by + c appear?
It is important to be able to look at an equation and understand what its graph will look like. You must be able to determine what calculation to perform on each x-value to produce its corresponding y-value.
For example, here is the graph of y = x^{2}.
The equation y = x^{2} tells you that for every x-value, you must square the x-value to find its corresponding y-value. Let's explore the graph with a few x-coordinates:
An x-value of 1 produces what y-value? Plug x = 1 into y = x^{2}. When x = 1, y = 1^{2}, so y = 1. So, you know a coordinate in the graph of y = x^{2} is (1,1).
An x-value of 2 produces what y-value? Plug x = 2 into y = x^{2}. When x = 2, y = 2^{2}, so y = 4. Therefore, you know a coordinate in the graph of y = x^{2} is (2,4).
An x-value of 3 produces what y-value? Plug x = 3 into y = x^{2}. When x = 3, y = 3^{2}, so y = 9. That determines that a coordinate in the graph of y = x^{2} is (3,9).
Tip: Solving the formula of a parabola for x tells you the x intercept (or intercepts) of the parabola—that is, where the parabola crosses the x-axis. If you get two real values for x, the parabola crosses the x-axis at two points. If you get one real root, then that value is the vertex. If both roots are complex, then the parabola never crosses the x-axis.
Comparing Graphs of Parabolas
Okay, now what if you are asked to compare the graph of y = x^{2} with the graph of y = (x – 1)^{2}? Let's compare what happens when you plug numbers (x-values) into y = (x – 1)^{2} with what happens when you plug numbers (x-values) into y = x^{2}:
y = x^{2} |
y = (x – 1)^{2} |
If x = 1, y = 1. |
If x = 1, y = 0. |
If x = 2, y = 4. |
If x = 2, y = 1. |
If x = 3, y = 9. |
If x = 3, y = 4. |
If x = 4, y = 16. |
If x = 4, y = 9. |
The two equations have the same y-values, but they match up with different x-values because y = (x – 1)^{2} subtracts 1 before squaring the x-value. As a result, the graph of y = (x – 1)^{2} looks identical to the graph of y = x^{2} except that the base is shifted to the right (on the x-axis) by 1:
How would the graph of y = x^{2} compare with the graph of y = x^{2} – 1?
In order to find a y-value with y = x^{2}, you square the x-value. In order to find a y-value with y = x^{2} – 1, square the x-value and then subtract 1. This means the graph of y = x^{2} – 1 looks identical to the graph of y = x^{2} except that the base is shifted down (on the y-axis) by 1:
Word Problems Aren't a Problem
You can easily solve the word problems using quadratic equations. Let's look carefully at an example.
Example
You have a patio that is 8 ft. by 10 ft. You want to increase the size of the patio to 168 square ft. by adding the same length to both sides of the patio. Let x = the length you will add to each side of the patio. You find the area of a rectangle by multiplying the length times the width. The new area of the patio will be 168 square ft.
(x + 8)(x + 10) = 168 |
FOIL the factors (x + 8)(x + 10). |
x^{2} + 10x + 8x + 80 = 168 |
Simplify. |
x^{2} + 18x + 80 = 168 |
Subtract 168 from both sides of the equation. |
x^{2} + 18x + 80 – 168 = 168 – 168 |
Simplify both sides of the equation. |
x^{2} + 18x – 88 = 0 |
Factor. |
(x + 22)(x – 4) = 0 |
Set each factor equal to zero. |
x + 22 = 0 and x – 4 = 0 |
Solve the equation. |
x + 22 = 0 |
Subtract 22 from both sides of the equation. |
x + 22 – 22 = 0 – 22 |
Simplify both sides of the equation. |
x = –22 |
Solve the equation. |
x – 4 = 0 |
Add 4 to both sides of the equation. |
x – 4 + 4 = 0 + 4 |
Simplify both sides of the equation. |
x = 4 |
Because this is a quadratic equation, you can expect two answers. The answers are 4 and –22. However, –22 is not a reasonable answer. You cannot have a negative length. Therefore, the only answer is 4.
To check your calculations, review the original dimensions of the patio—8 ft. by 10 ft. If you were to add 4 to each side, the new dimensions would be 12 ft. by 14 ft. When you multiply 12 times 14, you get 168 square ft., which is the new area you wanted.
What is the Quadratic Formula?
You can use the quadratic formula to solve quadratic equations. You might be asking yourself, "Why do I need to learn another method for solving quadratic equations when I already know how to solve them by using factoring?" Well, not all quadratic equations can be solved using factoring. You use the factoring method because it is faster and easier, but it will not always work. However, the quadratic formula will always work.
The quadratic formula is a formula that allows you to solve any quadratic equation—no matter how simple or difficult. If the equation is written in the form ax^{2} + bx + c = 0, then the two solutions for x will be x = . It is the ± in the formula that gives us the two answers: one with + in that spot, and one with –. The formula contains a radical, which is one of the reasons you studied radicals in the previous lesson. To use the formula, you substitute the values of a, b, and c into the formula and then carry out the calculations.
Example
Tip: To use the quadratic formula, you need to know the a, b, and c of the equation. However, before you can determine what a, b, and c are, the equation must be in ax^{2}+ bx + c = 0 form. For example, the equation 5x^{2} + 2x = 9 must be transformed to ax^{2}+ bx + c = 0 form.
Solving Quadratic Equations that have a Radical in the Answer
Some equations will have radicals in their answers. The strategy for solving these equations is the same as the equations you just completed.
Example
Find practice problems and solutions for these concepts at Quadratic Trinomials, Quadratic Equations, and the Quadratic Formula Practice Problems.
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