## Tuesday, June 19, 2012

### Problem of the Week Day 2: Week of 6/18/12 - 6/22/12

It is now day two of June’s problem of the week. Good luck!
Easy: Take out your answers to h and a, and solve the problem below:
a^5 + {a + √[-(a x h - h^3 + h ÷ a)]} = z
It looks straight forward, but there is a catch. You cannot solve these problems left to right, or in the order that is most convenient to you. There is a specific way of solving these problems, called the order of operations.
You must solve the problem in this order:
1. Solve what is inside any Parentheses or brackets
2. Solve any Exponents or roots
3. From left to right, complete all Multiplication and Division
4. From left to right, complete all Addition and Subtraction
You can remember this order with the phrase “Please Excuse My Dear Aunt Sally,” standing for parentheses, exponents, multiplication, division, addition, subtraction.
For example, look at this problem:
8 + 2^(2 x 2) ÷ 4
In this case, you would have to do the following:
8 + 2^(2 x 2) ÷ 4
8 + 2^4 ÷ 4
8 + 16 ÷ 4
8 + 4
12
It seems a little weird, but it is the correct way of solving it. Good luck. Remember to jot down z as well.
Hard: Now that we finished our trigonometry day, let’s move on to some Algebra. This day will probably be the longest day of the week for the hard problem, so get ready.
(f, 1808.51)
(g, 14)
(h, -6)
If you plot these three points on the Cartesian Plane, what equation goes through all of these points?
To solve this, you use something called systems of linear equations. Say we had to do it with the points (3, 0), (1, -8), and (-2, -5).
First off, you need to figure out how many points there are, and subtract one. This is the degree we are working with. In this case, we have three points, so we have a quadratic equation, or an equation with an x^2 in it.
Next, take the equation y = ax^2 + bx + c (for quadratic). We know some values that will come out for y, right? The three points above! So, we will plug all of those in for x.
0 = a(3)^2 + b(3) + c
0 = 9a + 3b + c
-8 = a(1)^2 + b(1) + c
-8 = a + b + c
-5 = a(-2)^2 + b(-2) + c
-5 = 4a - 2b + c
Now, we will solve for a, b, and c. To do this, we first have to eliminate a variable so it is just two variables. Conveniently, all of the c’s have the same coefficient. So, we will just subtract all of the equations from each other.
(9a + 3b + c) - (a + b + c) = 0 - -8
9a + 3b + c - a - b - c = 8
(9a - a) + (3b - b) + (c - c) = 8
8a + 2b + 0 = 8
8a + 2b = 8
(4a - 2b + c) - (a + b + c) = -5 - -8
4a - 2b + c - a - b - c = 3
(4a - a) + (-2b - b) + (c - c) = 3
3a - 3b + 0 = 3
3a - 3b = 3
With two variables, you only need two equations. Now, we must eliminate another variable. To do this, we must create a common coefficient. Since there is none, we will multiply both equations by something to do so.
3(8a + 2b) = 3(8)
24a + 6b = 24
-2(3a - 3b) = -2(3)
-6a + 6b = -6
Now, we will subtract just like before.
(24a + 6b) - (-6a + 6b) = 24 - -6
24a + 6b + 6a - 6b = 30
(24a + 6a) + (6b - 6b) = 30
30a + 0 = 30
30a = 30
And now, we have an equation we can solve.
30a = 30
a = 1
Since we have found a, we can plug that back into one of the two variable equations to get b.
3a - 3b = 3
3(1) - 3b = 3
3 - 3b = 3
-3b - 0
b = 0
Since we now have b, we can plug that into one of the original equations to get c.
-8 = (1) + (0) + c
-8 = 1 + c
-9 = c
So, we have:
a = 1
b = 0
c = -9
If we plug that into our y = ax^2 + bx + c, we get:
y = 1x^2 + 0x + -9
y = x^2 - 9
And there is your equation. Go through the same exact process as this, but with the other points. Record down a, b, c, and the equation you came up with. You will need it tomorrow.