## Wednesday, June 20, 2012

### Problem of the Week Day 3: Week of 6/18/12 - 6/22/12

Today is day three of the problem of the week! Good luck!
Easy: Today, we will be solving for a letter, but there is one difference. The letter is inside of the equation!
zn - h^2 = (z + h + a)n
In this problem, we are trying to figure out what n equals. Let’s do an example.
4n - 3 = 7n - 6
This problem is a little bit like a seesaw, in that if you take a certain amount off of one side, you take the same amount off the other side to keep it balanced. Also, if you add a certain amount to one side, you add the same amount to the other side. If you multiply or divide by something on one side, you do the same to the other side.
The first thing you have to do is put all of the variables (letters) on one side and all of the constants (numbers) on the other side. The first thing I would to is take 4n away from both sides.
4n - 3 = 7n - 6
4n - 3 - 4n = 7n - 6 - 4n
(4n - 4n) - 3 = (7n - 4n) - 6
0 - 3 = 3n - 6
-3 = 3n - 6
We have just made it so there is only one n. This makes things much simpler. Since all of the variables are on the right, we want all of the constants on the left. To do that, we will add six to both sides.
-3 = 3n - 6
-3 + 6 = 3n + (- 6 + 6)
3 = 3n + 0
3 = 3n
We have one step left. All we have to do is divide both sides by three.
3 = 3n
3 ÷ 3 = 3n ÷ 3
1 = n
And we have solved it. n = 1. You will do the same thing with the equation up top.
Hard:
Yesterday, you solved a system and got a quadratic equation. Today, you are going to solve the equation for x.
0 = ax^2 + bx + c
x1 = ___
x2 = ___
Please make x1 the smaller number and x2 the larger number. I will explain why there is an x1 and x2 in just a minute. First, let me explain how to solve the equation.
You could solve this equation with the quadratic formula, which I talked about a few times on this blog. You could also use techniques like completing the square, graphing, or the one I am going to talk about now: factoring.
Let’s say you have to solve the equation 0 = 2x^2 + 11x + 15. First, you have to do something called product sum. As it suggests, you need a product and a sum. The product is a x c. The sum is just b.
Product: 2 x 15 = 30
Sum: 11
Now, you need to find two numbers who have a product of 30 and a sum of 11. In this case, you can use 6 and 5.
6 x 5 = 30
6 + 5 = 11
Now, you need to do something called regrouping. For this, you will break the eleven up into the six and five.
0 = 2x^2 + 11x + 15
0 = 2x^2 + 6x + 5x + 15
Now, group together the first two terms and group together the second two terms.
0 = 2x^2 + 6x + 5x + 15
0 = (2x^2 + 6x) + (5x + 15)
Next, you will find the greatest common factor in each one. This means that for the 2x^2 + 6x, you will find a number that divides into both of those numbers. You will then factor it out.
0 = 2x(x + 3) + 5(x + 3)
Then, you will pull out the common binomial, which is x + 3. You are then left with 2x + 5.
0 = (x + 3)(2x + 5)
Since these are equal to zero, you know that one of the factors must equal zero. If neither equaled zero, then they couldn’t multiply together to equal zero. This means:
x + 3 = 0
x = -3
2x + 5 = 0
2x = -5
x = -5/2
So, x = -3 and x = -5/2. There are two solutions, which is fine. That is why there is an x1 and an x2.
Let’s try one more, just in case you are a little confused. Try 0 = x^2 + 12x + 35. First, we will find the product and sum.
Product: 1 x 35 = 35
Sum: 12
Now, we will find two numbers that have a product of 35 and a sum of 12.
7 x 5 = 35
7 + 5 = 12
Now, we will regroup by breaking the 12 into 7 and 5.
0 = x^2 + 5x + 7x + 35
Next, we will group the first two terms and the second two terms.
0 = (x^2 + 5x) + (7x + 35)
Now, we will find the greatest common factor.
0 = x(x + 5) + 7(x + 5)
Then, we will take the common binomial, and have it multiplied by what’s left.
0 = (x + 5)(x + 7)
And finally, we can solve it.
x + 5 = 0
x = -5
x + 7 = 0
x = -7
Good luck!