Learning about quadratic functions is similar to learning English: several words convey the same meaning. For example, demonstrate, illustrate, convey, depict, and portray are words that describe the act of showing.

This is similar to describing an important facet of quadratic functions: where y equals zero. Solution, solution set, x-intercepts, zeros, and roots all describe where y equals zero in a quadratic function.

1. Solution

Definition: The solution is the value of x that makes the function true.

What is the solution of this quadratic function?

0 = x2

1. Solve for x by taking the square root of 0 and x2.

Square root of 0 = 0
Square root of x2= x
0 = x

The solution is 0.

2. Plug in 0 for x to verify the answer.

0 = x2
0 = 02
0 = 0

2. Solution Set

Definition: The solution set includes 2 values of the input that make the function true.

What is the solution set of the following function?

k2 + 11k = -28

1. Set the function equal to 0.

k2 + 11k + 28 = -28 +28
k2 + 11k + 28 = 0

2. Solve for k by factoring.

k2 + 11k + 28 = 0

(k + 4)(k+7) = 0

k + 4 = 0
k = -4

k + 7 = 0
k = -7

Solution set: -4, -7

Is the solution set true?

1. Test k = -4 by plugging it in to the original equation.

k2 + 11k = -28
(-4)2 + 11(-4) = -28
16 + -44 = -28
-28 = -28

k = -4 is true.

2. Test k= -7 by plugging it in to the original equation.

k2 + 11k = -28
(-7)2 + 11(-7) = -28
49 + -77 = -28
-28 = -28

3. x-intercepts

Definition: An x-intercept is where a function and the x-axis intersect.

Usually, x-intercepts are used in the context of graphing, as in "Look at the graph and find the x-intercepts."

4. Zeros/Roots

Definition: A zero, or root, describes the input of a function when the output is 0. A quadratic function will have 0, 1, or 2 zeros or roots.

Find the zero(s) or root(s) or the following function:

y = 3a2 -10a + 8

1. Set the function equal to 0. Replace y with 0.

0 = 3a2 – 10a + 8

2. Solve by factoring.

0 = (3a + -4)(a + -2)

Set the factors equal to 0 and solve for a.

3a + - 4 = 0
3a + -4 + 4 = 0 + 4
3a = 4
3a/3 = 4/3
a = 4/3

a + -2 + 2 = 0 + 2
a + -2= 0
a = 2

3. Verify that when a = 4/3, y = 0

0 = 3a2 – 10a + 8
0 = 3(4/3)2 -10(4/3) + 8
0 = -24/3 + 8
0 = 0

4. Verify that when a=2, y = 0

0= 3a2 – 10a + 8
0= 3(2)2 - 10(2) + 8
0= 12 – 20 + 8
0 = 0

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