Solving Quadratic Equations

Here are a few sites that may help you when solving quadratic equations:

• by factoring
• by finding square roots
• by completing the square
• by Quadratic Formula
• by graphing
• extra examples

Solving Systems of Equations Using Elimination

I have found several websites that might help you with this concept.

Purple Math has a great site that shows examples here.

Check out the Unusual Outcomes in this PDF file. The other information there is also very good. The explanations are detailed.

Use this PDF file to help you determine when to add and when to multiply and then add.

Here are some problems that you can try to determine if you are getting it. Their answers are provided as well.

Polynomial Operations and Factoring

Here are some websites that might help clarify polynomial operations and factoring.

Polynomial addition and subtraction - http://math.about.com/library/blpoly.htm

Adding polynomials - http://www.regentsprep.org/rEGENTS/math/polyadd/sp_add.htm

Subtracting polynomials - http://www.regentsprep.org/rEGENTS/math/polyadd/sp_subt.htm

Multiplying polynomials - http://www.purplemath.com/modules/polymult.htm and http://www.jamesbrennan.org/algebra/polynomials/multiplication_of_polynomials.htm

Factoring polynomials - http://www.jamesbrennan.org/algebra/polynomials/factoring_polynomials.htm and http://www.purplemath.com/modules/simpfact.htm

In order for my students to earn extra credit, they must visit at least 2 of the sites and comment on both.

Pythagorean Theorem

The Pythagorean Theorem is used to find a missing side length of a right triangle. It only works for right triangles (one with a right angle).

The formula to use is a² + b² = c². A really good explanation of the formula and its derivation can be found here.

In order to use the theorem, you must know the parts of the triangle. Parts a and b of the theorem are the legs of the triangle. These are the side lengths that make up the right angle. The third side (longest side) is noted as c in the theorem.

Now substitute the values given into the theorem (formula) and solve the equation.

Example: A triangle has a side length of 6 cm and a side length of 8 cm. What is the length of the hypotenuse?

6² + 8² = c²
36 + 64 = c²
100 = c²
√100 = √c²
10 = c

Example: A triangle has a side length of 6 cm and a hypotenuse length of 12 cm. What is the length of the missing side?

a² + 6² = 12²
a² + 36 = 144
a² + 36 – 36 = 144 – 36
a² = 108
√a² = √108
a = √108
a = 6√3

Picture of triangle above from http://en.wikibooks.org/wiki/File:Right_triangle_shows_hyp_legs.PNG

Rate-Time-Distance Word Problems Part 2

Now for the second part of the rate-time-distance problems. This example refers to opposite direction travel.

Let’s look at the following example:

Train 1 leaves the train station travelling south at a rate of speed that is 10 mph more than train 2 which is travelling north. After 3 hours, they are 450 miles apart. Find the rate of train 2.

We are going to use another table to organize the rate, time and distance information.

Train	Rate	Time	Distance
1	r +10	3	3 (r + 10)
2	r	3	3r

Since the trains are moving in opposite directions, we are going to add their distances together to get 450 miles.

3 (r + 10) + 3r = 450
3r + 30 + 3r = 450
6r + 30 = 450
6r = 420
r = 70

Therefore the rate of train 2 is 70 mph.

Rate-Time-Distance Word Problems

Or otherwise known as those dreaded train problems. These are always tricky for algebra students. Ask someone you know which problems they remember from Algebra I and almost all of them will have something to say about the train problems.

Train problems can be separated into two categories: same direction travel and opposite direction travel. Each are handled differently, but using a chart makes it easier to set up the equations. You also have to remember that distance equals rate times time or d=rt.

Here is an example. A train leaves the train station at 2:00 p.m. Its average rate of speed is 90 mph. Another train leaves the same station a half hour later. Its average rate of speed is 120 mph. If the second train follows the same route on a parallel track to the first, how many hours will it take the second train to catch the first?

Train	Rate	Time	Distance
1	90	t	90t
2	120	t-0.5	120(t-0.5)

Since the trains are travelling in the same direction, their distances are equal when the 2nd one catches the 1st. Therefore, to solve this equation, we set the distance of Train 1 equal to Train 2.

90t = 120(t-0.5)
90t = 120t - 60
-30t = -60
t = 2

The first train travelled for 2 hours before the 2nd train caught up to it. The problem asks how long it takes the 2nd train to catch the first. So it takes the 2nd train a half hour less than it did the first train which is 1.5 hours.

Check back tomorrow for information on opposite direction travel.

Slope Y-Intercept Form to Standard Form

My students are having some trouble with changing slope y-intercept form of a line to standard form and vice versa. You must first understand solving equations before this will make sense. Having a good grasp of variables on both sides of the equation will help since it involves moving a variable term from one side of the equation to the other.

Here is an example:

Rewrite y = 2x - 5 (slope y-intercept form) as standard form.

Move the 2x to the left side of the equation with the y. -2x + y = -5

That is all that is required for this one. It is now in standard form (Ax + By = C).

Another part of the assignment that we worked on required that the standard form be written with integers (no fractions).

Rewrite 3/5y = 2/3x + 2 in standard form using integers.

Move the x term to the left side of the equation. -2/3x + 3/5y = 2
Now to get rid of the fractions, you will multiply both sides of the equation by 15 (the least common multiple of 3 and 5). You could multiply by any multiple of 3 and 5 but generally we use the least common multiple. As long as this multiplication is done to BOTH sides, the equation will stay balanced. Hint: Remember that you must use the distributive property on the left side of the equation.

15[-2/3x + 3/5y] = 2(15)

-30/3x + 45/5y = 30

-10x + 9y = 30 (standard form using integers)