Question 1:
Solve the following quadratic equation:
5x2 + 20x – 466 = 119
The General Form of a quadratic is y = ax2 + bx + c
Step 1: Set the
equation equal to zero. So in this case, subtract 119 from each side.
5x2
+ 20x – 466 = 119
-119 -119
5x2 + 20x – 585 = 0
Step 2: Factor
out the greatest common factor, or the largest number that goes into each term.
For this equation, it is 5.
5x2
+ 20x – 585 = 0
5 5 5
5(x2 + 4x – 117) = 0
If you didn’t realize it factored, follow this link: http://www.youtube.com/watch?v=1lmO0Np1Jbw
Note: After factoring out
the GCF, it always stays with the equation outside the parenthesis.
Step 3: Determine
what factors multiply to get -117 that also add up to 4.
Factors
that multiply to get -117: 3*-39, -3*39, -13*9 and 13*-9
Of these,
only 13*-9 add up to get 4.
Step 4: Convert
the equation to factored form.
5(x + 13)(x
– 9) = 0
You know
this works because if you expand it, you get the equation set equal to zero
which was: 5x2 + 20x – 585 = 0.
Step 5: Solve for
x. To solve for x, you must set both (x + 13) and (x – 9) equal to zero then
get x all by its self for both.
(x + 13) = 0 (x
– 9) = 0
-13 -13 +9 +9
x = -13 x
= 9
Now that we have factored the equation and solved for x,
all we have left to do is find the vertex of the equation.
Find the vertex:
5x2 + 20x – 585 = 0
Step 1: Find the “x” of the vertex. The “x” of the vertex equals –b
(2 * a)
x = -20 x
= -20 x = -2
(2 * 5)
10
So the x of the vertex = -2
Note: In this
case the “-b” means opposite b. If the b value was negative, the number would
become positive and vice versa.
Step 2: Plug -2 back into the original equation to get the “y” of
the vertex.
y = 5(-2)2 + 20(-2) – 585 y = 20 – 40 – 585 y
= -605
Vertex: (-2, -605)
Graph the equation
5x2 + 20x – 585 = 0
Step 1: The only value we don’t have is the y – intercept. The y –
intercept is the constant number, or the number that doesn’t change, in each
equation. The only number that will not change is -585.
The x – intercepts we found when solving the original
equation for x.
X – Intercepts: (-13, 0) (9, 0)
Y – Intercept: (0, -585)
Vertex: (-2, -605)
Solve the following equation:
Log21(4)
+ Log21(3x+2) = Log21(4x+1) - Log21(9x+2) + 2
Step 1: Get all the logs
on onto one side.
Log21(4) + Log21(3x+2) = Log21(4x+1)
- Log21(9x+2) + 2
-Log21(4x+1) - Log21(4x+1)
+ Log21(9x+2)
The equation now looks like:
Log21(4) + Log21(3x+2)) + Log21(9x+2) - Log21(4x+1 = 2
Step 2: There are 8 rules
of logarithms, two of which apply here.
The product rule of logarithms: LogaM + LogaN
= LogaMN
The quotient rule of logarithms: LogaM - LogaN
= LogM
N
Applying these rules, the new equation is:
Log21(4)(3x+2)(9x+2) = 2
(4x+1)
Step 3: Turn the equation
into an exponential.
212 = (4)(3x+2)(9x+2) 441 = (4)(3x+2)(9x+2)
(4x+1)
(4x+1)
Step 4: Multiply both
sides by the denominator.
(4x+1) * 441 = (4)(3x+2)(9x+2) * (4x+1) = 1764x
+ 441 = 108x2 + 96x + 16
(4x+1)
Step 5: Set the equation
equal to zero.
1764x + 441 = 108x2 + 96x + 16
-1764x -441 -1764x -441
108x2 – 1168x – 380 = 0
Step 6: Complete the
square. Add the constant to the other side.
108x2 – 1168x – 380 = 0
+ 380 + 380
108x2 – 1168x = 380
Step 7: Factor out the “a”
value.
108(x2 – 1168x) = 380
108
Step 8: Find the perfect “c”
value using (b2) and then add a * (b2) to
the other side.
2 2
108(x2 – 1168x + 21316) = 380 +
(108 * 21316 )
108
729 729
Step 9: Factor into the
perfect square binomial.
108(x – 1168)2 = 1768962963
216
500000
Step 10: Divide each side
by the “a” value.
108(x – 1168)2 = 1768962963
216 500000
108
108
Step 11: Take the square
root of each side.
√(x – 1168)2 = √3275857337
100000000
Step 12: Isolate the
variable.
x – 1168 = +-5723510581 + 1168
+ 1168 1000000000
Step 13: Solve for x.
x = 5723510581 + 1168 = 1173723511 x =
-5723510581 + 1168 = 1162276489
1000000000 1000000 1000000000 1000000
Question 3
Factor the following polynomial and solve:
7x4
+ 35x3 – 84x2 – 420x = 0
Step 1: Factor out the
greatest common factor.
7x(x3 + 5x2 – 12x – 60) = 0
Step 2: Put the equation
into two sets of parenthesis. In order to do this, two things must be true:
1: The four terms in each
equation must be increasing by the same increment. In this case, 5.
2: All numbers must have
the same sign.
7x(x3 + 5x2) – (12x + 60) = 0
Note: In the
second set of parenthesis, it becomes positive 60 instead of negative 60. This is
because subtracting a negative is the same thing as adding a positive value. So
we change the sign to accommodate rule #2.
Step 3: Factor out the
greatest common factor from each parenthesis.
7x[x2(x + 5) – 12(x + 5) = 0
Step 4: Combine the
greatest common factor from both parenthesis into one.
7x(x2 – 12)(x + 5) = 0
Step 5: Solve for x.
7x = 0 x = 0 x + 5 = 0 x = -5
7 7 - 5 - 5
x2 – 12 = 0 √x2
= √12 x = +-√12
+ 12 + 12
Question 4
Find the domain of the following equation:
√3x2
+21x + 36 = 0
72x + 9327
Note: The entire numerator of the equation is under the radical. My
computer just doesn’t allow me to do that.
Step 1: Let’s just start
with the numerator to begin with. Factor out the greatest common factor.
3(x2 + 7x + 12) = 0
Step 2: Factor the
numerator.
3(x + 4)(x + 3)
Step 3: Determine the
range for the numerator by solving for x. Because the numerator is under a radical,
the equation cannot be a negative number. So we must set the terms greater than
or equal to zero.
x + 4 ≥ 0 x
+ 3 ≥ 0
- 4 - 4 - 3 - 3
x ≥ -4 x
≥ -3
Step 4: Now let’s find
the range for the denominator by solving for x. However, because this is in the
denominator, the term cannot equal zero because dividing by zero is undefinable.
So x cannot equal zero.
72x + 9327 ≠ 0
- 9327
- 9327
72x ≠ -9327
72 72
x ≠ -129.542
Step 5: Combine both
domains into one.
(-∞, -129.542) U [-4, -3] U [-3, ∞)
You may notice there are parentheses
around certain numbers and brackets around others. Because (-)∞ is not
reachable and cannot be plugged into an equation, a parenthesis is used. One is
also used on -129.542 because the number cannot be plugged into the denominator
or else it would equal zero, which isn’t allowed. Brackets are used on the
other numbers because they can be plugged into the equations. This is because
under a radical, the number cannot be negative. Plugging the numbers in would
not make any negative. The U means union which just combines the domains together.
Reflection:
This project wasn't fun for me like I thought it would be. It turned out to just be a whole lot of stress. I thought I was going to be able to do something really creative and fun, but, because of how crappy my computer is and my lack of internet, I had to do a crappy extended blog post. I picked the problems that I did because they were what I struggled with from the respective units. I thought that creating problems and solving them would help me get a better grasp on the material, which it did. Making the problems was a struggle in its self. I hated having to be able to come up with equations that were not only challenging but solvable. Running through everything step by step for the post made things seem to make a lot more sense. That’s why I chose the problems that I did. These problems provide my best mathematical understanding because I tried to make them very challenging. For me, each problem took about an hour to two hours to create to make them complicated. These are all hard question for me so to be able to create them and solve them is a great way to prove my understanding of what I’ve learned thus far. This was somewhat valuable to me. I don’t think it helped as much as it could. I still need to use my notes to complete some types of problems such as completing the square. However, it did make me have a better overview of the subjects that I covered. I learned that hard-work pays off for projects like this. Spreading my time out more wisely would have benefitted me more for this project. I didn’t like how broad this project was. If the questions had designated topics, it would have been easier for me to come up with the problems. Leaving it as broad as it was freaked me out and put me in a “deer in headlights” type trance that made me get lost trying to figure out what to do.
Reflection:
This project wasn't fun for me like I thought it would be. It turned out to just be a whole lot of stress. I thought I was going to be able to do something really creative and fun, but, because of how crappy my computer is and my lack of internet, I had to do a crappy extended blog post. I picked the problems that I did because they were what I struggled with from the respective units. I thought that creating problems and solving them would help me get a better grasp on the material, which it did. Making the problems was a struggle in its self. I hated having to be able to come up with equations that were not only challenging but solvable. Running through everything step by step for the post made things seem to make a lot more sense. That’s why I chose the problems that I did. These problems provide my best mathematical understanding because I tried to make them very challenging. For me, each problem took about an hour to two hours to create to make them complicated. These are all hard question for me so to be able to create them and solve them is a great way to prove my understanding of what I’ve learned thus far. This was somewhat valuable to me. I don’t think it helped as much as it could. I still need to use my notes to complete some types of problems such as completing the square. However, it did make me have a better overview of the subjects that I covered. I learned that hard-work pays off for projects like this. Spreading my time out more wisely would have benefitted me more for this project. I didn’t like how broad this project was. If the questions had designated topics, it would have been easier for me to come up with the problems. Leaving it as broad as it was freaked me out and put me in a “deer in headlights” type trance that made me get lost trying to figure out what to do.
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