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Any questions on the Section 5.8 homework?
Pass your worksheets for this assignment to the middle
aisle for pickup now.
Remember the problem like this one from the homework that was due today?
Wouldn’t it be nice if there was an easier way to do it than by factoring?
Leave factoring up on board: (4x - 9)(3x + 8)
Please
CLOSE
YOUR LAPTOPSand turn off and put
away your cell phones.Sample
Problems Page Link
(Dr. Bruce Johnston)
Section 8.2
The Quadratic Formula
The Quadratic Formula
The quadratic formula is another technique we can use for solving quadratic equations.
Remember, quadratic equations are polynomial equations of degree 2, such as
x2 + 3x -7 = 0
or
5x2 – 14 = 0.
The quadratic formula is derived from a process called “completing the square” for a general quadratic equation. – See Section 8.1 if you’re interested in seeing how
this formula is derived. – This will also be covered in Math 120 in more
detail, along with the technique called “completing the square”.
The Quadratic Formula:
a
acbbx
2
42
The solutions to the equation ax2 + bx + c = 0
are given by the formula
Note: This formula IS on the pink formula sheet, but you’ll probably have it memorized by the time you’ve done the first few homework problems.
The Big Question:
How can we tell when we should use
factoring and when we should use the quadratic formula?
Solve x2 + 4x + 3 = 0 by
• Factoring• The quadratic formula.
Which way works best?
Example 1
Solve x2 + 4x + 3 = 0 by Factoring:
This one is pretty easy to factor. The factoring is (x + 3)(x + 1) = 0, so the solutions are given by x + 3 = 0, or x = -3,
and x + 1 = 0, which gives x = -1.
Now, solve x2 + 4x + 3 = 0 by the quadratic formula:
a = 1, b = 4, c = 3, so the formula gives:
32
6
2
24
12
2
2
24
2
24
2
44
2
12164
12
31444 2
or
x
Which way works best for this problem?
In this case, the factoring method is much quicker, although BOTH methods give the same answer.
Solve x2 + 5x + 12 = 0 by
• Factoring• The quadratic formula.
Which way works best?
Example 2
Solve x2 + 5x + 12 = 0 by Factoring: This one looks pretty easy to factor, but when you
start trying to find two factors of 12 that add up to 5, nothing works.
(1+12=13, 2+6=8, 3+4=7).
What does this mean? It means that the polynomial is PRIME, and there
are no rational solutions. (Remember, a rational number is either an integer or a fraction.)
Solve x2 + 5x + 12 = 0 (continued):
• Let’s see what the quadratic formula gives in this case: a = 1, b = 5, c = 12
so the formula gives:
2
235
2
48255
12
121455 2
x
Notice that the number under the radical sign is negative, which means there are no real answers. If the number under the square root sign comes out to be positive but it’s not a perfect square, this means the answer is a real number, but is irrational because it can’t be simplified to remove the radical. In either of these cases, we’d say the polynomial is prime, and therefore has no rational roots.
So which way works best for solving
x2 + 5x + 12 = 0?
Either way works fine, but if you think a polynomial is prime, a good way to check is by calculating the discriminant (b2 – 4ac). If the discriminant is either negative or not a perfect square, then you know for sure that your polynomial is prime and there are no rational solutions.
Now re-do this problem from the 5.8 homework using the quadratic formula:
Answers: -8/3, 9/4
Which way works best in this case?
Either way works, but the quadratic formula approach is probably going to be faster than factoring for most people.
Moral of the story: For a quadratic equation with a leading coefficient other than 1, it’s probably going to be quicker to solve it using the quadratic formula than it would be to factor the polynomial.
)1(2
)20)(1(4)8(8 2
x
2
80648
2
1448
2
128210,
2
4
2
20 or or
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = -20
8
1
2
5Solve x2 + x – = 0 by the quadratic formula.
Question: What if some coefficients in your quadratic equation are fractions? ANSWER: Clear them first by multiplying all terms by the LCD:
• The expression under the radical sign in the quadratic formula (b2 – 4ac) is called the discriminant.
• The discriminant will take on a value that is positive, 0, or negative.
• The value of the discriminant indicates two distinct real solutions (if it’s positive), one real solution (if it’s zero), or two complex, but not real solutions (if it’s negative – a topic to be discussed in Math 120).
No x-intercepts
No real solution; two complex imaginary solutions
b2 – 4ac < 0
One x-intercept
One real solution (a repeated solution)(If b2 – 4ac is a perfect square, the solution will be a rational number. If not, it’s irrational.)
b2 – 4ac = 0
Two x-intercepts
Two unequal real solutions(If b2 – 4ac is a perfect square, the two solutions will be rational numbers. If not, they’re both irrational.)
b2 – 4ac > 0
Graph of y = ax2 + bx + c
Kinds of solutions to ax2 + bx + c = 0
Discriminantb2 – 4ac
The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0
Use the discriminant to determine the number and type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = -4, and c = 5
b2 – 4ac = (-4)2 – 4(12)(5)
= 16 – 240
= -224
Since the discriminant is negative, there are no real solutions.
Question: What would this graph look like?
Example
Use the discriminant to determine the number and type of solutions for the following equation.
25x2 - 4 = 0
a = 25, b = 0 (why?) , and c = -4
b2 – 4ac = (0)2 – 4(25)(-4) = 0 – -400 = 400
Since the discriminant is positive, there are two real solutions.
(You could go on to show that the solutions are 2/5 and -2/5, either by factoring or using the quadratic formula.)
Example
Use the discriminant to determine the number and type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = -4, and c = 5
b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 24 = -224
Since the discriminant is negative, there are no real solutions.
Example
Use the discriminant to determine the number and type of solutions for the following equation.
x2 – 8x + 16 = 0
a = 1, b = -8, and c = 16
b2 – 4ac = (-8)2 – 4(1)(16) = 64 – 64 = 0
Since the discriminant is zero, there is one real solution. (You could go on to show that the solution is 4, either by factoring or using the quadratic formula.)
Question: What would this graph look like?
Example
2 possible approaches:
1. Exact answer: The exact answer will contain a radical, i.e. it will be an irrational number. (More on this in Chapter 7...)
2. Approximate answer: Use your calculator to get an approximate decimal answer.
How do you figure out the answers if the discriminant is positive but not a perfect square?
REMINDER!!!IMPORTANT NOTE:
Use the quadratic formula technique to solve all problems in this homework assignment.
There are a couple of word problems at the end of the assignment in which the online learning aids will show factoring as the solution method. You should use the quadratic formula instead (and you will find it to be easier and quicker than factoring.)
Reminder:
This homework assignment on Section 8.2 is due
at the start of next class period.
You may now OPEN
your LAPTOPSand begin working on the
homework assignment (if there’s any time left...)
But remember, you can always work in the JHSW 203 open lab after class (or before
tour next class session) if you want some help on this homework.