| Slide 23 |
Solving Quadratic Equations: Example
Hint: Before getting tangled up in arithmetic , think about what a
reasonable solution might be. Recall:
| Dose |
Weight Gain (dekagrams |
 |
If the animal gained 5 dekagrams, the dose was
probably between 6
and 7. |
| Slide 24 |
Solving Quadratic Equations: Example
Hint: When possible, draw a picture .
Dose |
| Slide 25 |
Solving Quadratic Equations: Example
If the animal had gained 5 dekagrams, substituting for y ,
y = 5 = 1.13 − 0.41x + 0.17x2
5 −5 = 1.13 − 5 − 0.41x + 0.17x2
0 = 0.17x2 − 0.41x − 3.87
We can use the quadratic equation with
a = 0.17, b = −0.41, c = −3.87 |
| Slide 26 |
Solving Quadratic Equations: Example
We should go back and check these! |
| Slide 27 |
Solving Quadratic Equations: Example
The solutions to the equation are 6.127 and -3.715. Do both of
these make sense in the context of our problem? No - animals
cannot receive a negative dose ! We are only interested in solutions
greater than zero . The predicted dose is 6.127. (We can envision a
dose bigger than 6, but less than 7.) This agrees with our best
estimate before doing the math . |
| Slide 28 |
Absolute Value Equations
Recall: Absolute value refers to a number’s distance from 0 on
the real number line.
|a − b| is the absolute value of (a − b), or the
distance from a to b
Example:
|5 − 3| = 2 |
| Slide 29 |
Absolute Value Equations
a + b| can be thought of as |a − (−b)|, or the
distance between a
and −b.
|5 + 3| = |5 − (−3)|, or the distance between 5 and -3, or 8.
| − a − b| = | − a − (+b)|, or the distance from −a to b
| − 5 − 3| = | − 5 − (+3)| = | − 8| = 8 |
| Slide 30 |
Solving Absolute Value Equations
|a| = 5 means, “What values of a are 5 units away
from 0?”
Solutions: 5 and -5
|a − 3| = 2 means, “What are the values of a such that the distance
between a and 3 is 2?”
We can deduce the answer by finding 3 on the number line and
finding the values 2 units away from 3.
Algebraically , we can solve the equations a −3 = 2 and a −3 = −2.
So a = 5 and a = 1 |
