1. Solving Quadratic Equations by the Quadratic Formula

2. Basics
A quadratic equation is an equation equivalent to an equation of the type
ax2 + bx + c = 0, where a is nonzero

3. WHY USE THE QUADRATIC FORMULA?
The quadratic formula allows you to solve ANY quadratic equation, even if you cannot factor it.
An important piece of the quadratic formula is what’s under the radical:
b2 – 4ac
This piece is called the discriminant.

4. WHAT THE DISCRIMINANT TELLS YOU!
Value of the Discriminant
Nature of the Solutions
Negative
2 imaginary solutions
Zero
1 Real Solution
Positive – perfect square
2 Reals- Rational
Positive – non-perfect square
2 Reals- Irrational

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7. Imaginary number i
Up until now, you've been told that you can't take the square root of a negative number.
A new number called "i", standing for "imaginary", can make it happen! Because "everybody knew" that i wasn't "real". (That's why you couldn't take the square root of a negative number before: you only had "real" numbers; that is, numbers without the "i" in them.) The imaginary is defined to be:

8. x2 – 6x + 5 = 0 n2 – 18n + 81 = 0 4y2 – 12y + 9 = 0 9m2 + 24m + 16 = 0
Find the discriminant then state the number of rational, irrational, and imaginary solutions.
x2 – 6x + 5 = 0
n2 – 18n + 81 = 0
4y2 – 12y + 9 = 0
9m2 + 24m + 16 = 0
–7q2 + 8q + 2 = 0

9. THE QUADRATIC FORMULA
When you solve using completing the square on the general formula you get:
This is the quadratic formula!
Just identify a, b, and c then substitute into the formula.

10. x2 + 2x - 3 = 0 Solve for x: x2 + 2x = 3 Solution:
This equation isn't in the proper form -- we first need to subtract 3 from each side so there's a 0 on the right:
x2 + 2x - 3 = 0
Now we can just use the quadratic formula to get our answers, given that a=1, b=2, c= -3:

12. To the tune of ....
Pop Goes the Weasel
Math Version:
x equals negative b
plus or minus the square root
of b squared minus 4ac
all over 2a
Actual Song:
All around the mulberry bush,
The monkey chased the weasel.
The monkey thought 'twas all in fun. 
Pop! goes the weasel.

13. Use the quadratic formula to solve for x
3x2 -x -6= 0

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Let’s try one with a complex root
2x2 + 2x + 5

15. 2x2 + 2x + 5= 0

16. Find the roots: x2 + 4x + 5 = 0


19. The complex roots in this example are
x = -2 + i and x = -2 - i. These roots are identical except for the "sign" separating the two terms. One root is -2 PLUS i and the other root is -2 MINUS i. This pattern will occur in every set of complex roots that you will encounter when solving a quadratic equation. Roots that possess this pattern are called complex conjugates (or conjugate pairs).

20. Application of Quadratic Equations
Word Problems
Instead of giving you the question directly to solve, the problem is presented in the form of a story.
Based on the story, you have to form the equation and solve it

21. Steps to solving a word problem
Identify the ‘unknown’ – Let it be represented by a letter (x, y, a etc…)
Form the quadratic equation using the given information
Solve the equation
Check if your solutions satisfy the problem.

22. The sum of two numbers is 27 and their product is 50. Find the numbers.
Let one number be x. Then the other number is 50/x.
x + 50/x = 27
X x => x = 27x
- 27x => x2 - 27x + 50 = 0
(x -25)(x -2) = 0
(x -25) = 0 or (x -2) = 0
x = 25 or x = 2.

23. The sum of two numbers is 27 and their product is 50. Find the numbers.
Let one number be x. Then the other number is 50/x.
x + 50/x = 27
X x => x = 27x
- 27x => x2 - 27x + 50 = 0
(x -25)(x -2) = 0
(x -25) = 0 or (x -2) = 0
x = 25 or x = 2.

24. A ball is thrown upwards from a rooftop, 80m above the ground
A ball is thrown upwards from a rooftop, 80m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, which is given by,
h = -16t2 + 64t + 80.
What is the height reached by the ball after 1 second?
What is the maximum height reached by the ball?
How long will it take before hitting the ground?

26. What is the height reached by the ball after 1 second?
h = -16t t + 80
h = -16* 1*1 + 64* = 128m
What is the maximum height reached by the ball?
Rearrange by the completing the square, we get:
h = -16[t2 - 4t - 5]
h = -16[(t - 2)2 - 9]
h = -16(t - 2)
When the height is maximum, t = 2; therefore, maximum height = 144m.

27. How long will it take before hitting the ground?
When the ball hits the ground, h = 0;
-16t2 + 64t + 80 = 0
Divide the equation by -16
t2 - 4t - 5 = 0
(t - 5)(t + 1) = 0
t = 5 or t = -1
The time cannot be negative; so, the time = 5 seconds.

28. An object is launched at 19. 6 meters per second (m/s) from a 58
An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t t , where s is in meters. When does the object strike the ground?

29. What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So I'm looking for the time when the height is s = 0. I'll set s equal to zero, and solve:
0 = –4.9t t 
0 = t2 – 4t – 12 
0 = (t – 6)(t + 2)
Then t = 6 or t = –2. The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at –2, but negative time won't work in this word problem.) So "t = –2" is an extraneous solution, and I'll ignore it. The object strikes the ground six seconds after launch.

30. The product of two consecutive negative integers is 1122
The product of two consecutive negative integers is What are the numbers?

31. Consecutive integers are one unit apart, so my numbers are n and n + 1
Consecutive integers are one unit apart, so my numbers are n and n + 1. Multiplying to get the product, I get:
n(n + 1) = 1122 
n2 + n = 1122 
n2 + n – 1122 = 0 
(n + 34)(n – 33) = 0
The solutions are n = –34 and n = 33. I need a negative value, so I'll ignore "n = 33" and 
take n = –34. Then the other number is n + 1 = (–34) + 1 = –33.
The two numbers are –33 and –34.

32. A garden measuring 12 meters by 16 meters is to have a pedestrian pathway installed all around it, increasing the total area to 285 square meters. What will be the width of the pathway?

35. The first thing I need to do is draw a picture
The first thing I need to do is draw a picture. Since I don't know how wide the path will be, I'll label the width as "x".
  Looking at my picture, I see that the total width will be
x x = x, and the total length will be 
x x = x.
  Then the new area is given by:   
(12 + 2x)(16 + 2x) = 285 
 x + 4x2 = 285 
4x2 + 56x – 93 = 0

36. Obviously the negative value won't work in this context, so I'll ignore it.
The width of the pathway will be 1.5 meters.