This Blog exists for the collective benefit of all algebra students. While the posts are specific to Mr. Chamberlain's class, any and all "algebra-ticians" are welcome. The more specific your question (including your own attempts to answer it) the better.
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Saturday, May 12, 2012
Unit 9 Review
I notice that the MathChamber Academy videos for Unit 9 haven't been
viewed too often. If you want to get a flavor for the Unit 9 Test, do
Problem Set 9A and watch the videos.
Well, we KNOW that a good candidate for complete the square should have a coefficient of ONE for the x^2 term. Gosh Golly Gee, 4 is the GCF of all of these terms... Whoop-de-doo!!
x^2 - 2x = 6
hmmm... now what 'c' value would make the terms on the left into a perfect square trinomial? Well, half of -2 is -1 and -1 squared is ONE, so, let's add ONE to both sides...
x^2 - 2x + 1 = 7 (x-1)^2 = 7 (now let's take the sqrt of both sides) x-1 = +/- sqrt(7) (now let's add ONE tbs) x = 1 +/- 2.65 x = -1.65, 3.65
#11) 100 sq ft of banner... length should be 15 ft longer than the width. RELATE DEFINE WRITE
RELATE The product of the length and width of the banner is equal to a specified square footage
DEFINE w is the width of the banner w+15 is the length of the banner 100 sq ft is the square area of the banner
WRITE
w(w+15)=100 w^2 + 15w = 100 w^2 + 15w - 100 = 0
BOX & DIAMOND (I assume you can do that) yields +20 and -5 as the two numbers.
(w+20)(w-5)=0
Solution: w= -20,5
hmmmm... a banner of length -20? THAT makes no sense... we call that an extraneous solution, thus the solution of 5 for the width is the only sensible answer.
The width of the banner is 5 ft and the length is 20 ft.
Well, it's not square-rootable (it has a "middle" term) Not (integer) factorable (the "box & diamond" failed) Not a good candidate for Complete the Square (because the "middle" term coefficient is odd) SO... that leaves the Quadratic Formula (sing it for me!)
#38 is asking for the number of real number solutions for the equation... YES, you MUST set the equation = 0 in order to achieve standard form BEFORE you find your a,b,&c values.
#38) x^2 + 7x - 10 = 3
Subtract 3 fbs to obtain:
x^2 + 7x - 13 = 0
use the discriminant (b^2-4ac) to answer the question
a=1 b=7 c=-13
(7)^2 - 4(1)(-13) 49 - (-52) 49 + 52 101
The discriminant is positive, therefore there are TWO real number solutions...
Nvr mind I got the graphing!!! BUT.... I'm stuck on #14 on the test with w to equal getting 15 and 6. I understand how to set up the quad equation and solve by factoring. But when I factored, I got w = 15 and -3 (which is extraneous). I know the correct answer is 15 and 6 but I don't know how to get it.
OK, so let's do #14 on from the first test. From your question, I sense that you are OK with the set up of
w(2w+3)=90 which can be simplified to
2w^2 + 3w = 90
Hopefully, that part makes sense to everyone as far as it is a formula for the area of the deck.
... before we can use the ZPP, we need to set the equation = zero, so...
which can be simplified to
2w^2 + 3w - 90 = 0 (still no magic, right?)
This CAN be factored OR we can use the Quadratic formula
If we factor, we end up with
(2w + 15)(x - 6) = 0
which presents us with the solutions:
w = -7.5, 6
Since the context of the problem DEMANDS a positive solution, the only USEABLE solution is 6, since a rectangle can't be measured with a negative length, therefore the solution of -7.5 is EXTRANEOUS (i.e. irrelevant).
So, w=6 means that the width equals 6, and we go back to the original problem to see that the length is 2w+3, or in this case 15.
SO, the solutions to the quadratic mathematical model are -7.5 & 6, of which 6 is the only one that makes sense in the context of the real life problem.
SO, the ANSWER to the question is that the width is 6 and the length is 15.
It might seem "nit-picky" but do you see how I am using the words SOLUTION and ANSWER in different ways.
We SOLVE math (model) problems... we ANSWER real-life questions, and we sometimes have to pick through the "weeds" of the SOLUTIONS to the mathematical model to find our ANSWERS.
Mr.C I'm, having trouble on #11 on the Chapter test. I said the length was w+15. My quad formula was w(w+15)-100=0. After I put it in the quad equation I got w=20 and the length would equal 35 which DOES NOT equal 100. I think the right answer is w=5 and L=20 but I don't know how to get that!! Oh and btw #14 makes sense now. Thanx!
OH MY GOSH!!! I made a really big mistake in setting up my quad formula and forgot to make the b negative, so the 15 was postive!! Opps, now it make sense!!!!
Mr.C I don't understand how to solve #21 on pg 617 on hw #10-1. I tried mapping the joggers path into a triagnle and plugging in the numbers for the Pythagorean Theorem, but it didn't come out to the right answer. Could you help???
Mr.C, I'm having trouble solving the word problem (#9) on Problem Set 9A. Could you help???
ReplyDeleteI added this problem to MathChamber Academy Unit 9
Deletehttp://mathchamberacademy.pbworks.com/w/page/53257325/Unit%209%20-%20Quadratic%20Functions%20and%20Equations
lmk if it helps...
#34 in the review made little sense anyone?
ReplyDelete4x^2 - 8x = 24
DeleteWell, we KNOW that a good candidate for complete the square should have a coefficient of ONE for the x^2 term. Gosh Golly Gee, 4 is the GCF of all of these terms... Whoop-de-doo!!
x^2 - 2x = 6
hmmm... now what 'c' value would make the terms on the left into a perfect square trinomial? Well, half of -2 is -1 and -1 squared is ONE, so, let's add ONE to both sides...
x^2 - 2x + 1 = 7
(x-1)^2 = 7 (now let's take the sqrt of both sides)
x-1 = +/- sqrt(7) (now let's add ONE tbs)
x = 1 +/- 2.65
x = -1.65, 3.65
Voila!!
i have a quick question on numbers 11, 24 and 25 on the chapter test. 25 is worded weird.
ReplyDelete#11) 100 sq ft of banner... length should be 15 ft longer than the width.
DeleteRELATE DEFINE WRITE
RELATE
The product of the length and width of the banner is equal to a specified square footage
DEFINE
w is the width of the banner
w+15 is the length of the banner
100 sq ft is the square area of the banner
WRITE
w(w+15)=100
w^2 + 15w = 100
w^2 + 15w - 100 = 0
BOX & DIAMOND (I assume you can do that) yields +20 and -5 as the two numbers.
(w+20)(w-5)=0
Solution: w= -20,5
hmmmm... a banner of length -20? THAT makes no sense... we call that an extraneous solution, thus the solution of 5 for the width is the only sensible answer.
The width of the banner is 5 ft and the length is 20 ft.
Voila!
#24 Write an equation for a parabola that has a maximum value and two x-intercepts.
Deletehmmmm... well, if it has a max value, it must open down and thus have a neg. coefficient for the x^2 term.
hmmmm again.... if it has x-intercepts, the max value MUST be above the x-axis.
How about y = -x^2 + 4 ?? Yup, that's ONE example... there are an infinite number of other examples, but the question just asked for ONE equation.
#25 We (SHOULD) know that the only quadratic trinominal equations that have one root are PERFECT SQUARE TRINOMIALS... SO, what value of k would make:
ReplyDeletekx^2 - 10x +25 = 0
a Perfect Square Trinomial??
k must equal ONE.
Sometimes the toughest questions are the easiest.
Mr. C.
For problem 4 on problem set 9A i got a different answer from everyone in class. Can you please explain it.
ReplyDeletePlease let me know if this helps:
Delete#4) Solve: x^2 + 3x - 5 = 0
Well, it's not square-rootable (it has a "middle" term)
Not (integer) factorable (the "box & diamond" failed)
Not a good candidate for Complete the Square (because the "middle" term coefficient is odd)
SO... that leaves the Quadratic Formula (sing it for me!)
a=1
b=3
c=-5
x = {-(3) +/- Sqrt[(3)^2 - 4(1)(-5)]}/[2(1)]
x= {-3 +/- sqrt[ 9 - (-20)]}/2
x = [-3 - sqrt(29)]/2 , [-3 + sqrt(29)]/2
x = -4.2, 1.2
can someone explain number 38 on page 605 to me. are you not supposed to make the equation equal to zero before using the quadratic formula
ReplyDelete#38 is asking for the number of real number solutions for the equation... YES, you MUST set the equation = 0 in order to achieve standard form BEFORE you find your a,b,&c values.
Delete#38) x^2 + 7x - 10 = 3
Subtract 3 fbs to obtain:
x^2 + 7x - 13 = 0
use the discriminant (b^2-4ac) to answer the question
a=1
b=7
c=-13
(7)^2 - 4(1)(-13)
49 - (-52)
49 + 52
101
The discriminant is positive, therefore there are TWO real number solutions...
Capeesh?
Mr. C I'm still confused on how to graph problem 6 on the test without setting up a table. Could you help???
ReplyDeleteNvr mind I got the graphing!!! BUT.... I'm stuck on #14 on the test with w to equal getting 15 and 6. I understand how to set up the quad equation and solve by factoring. But when I factored, I got w = 15 and -3 (which is extraneous). I know the correct answer is 15 and 6 but I don't know how to get it.
ReplyDeleteOK, so let's do #14 on from the first test. From your question, I sense that you are OK with the set up of
Deletew(2w+3)=90
which can be simplified to
2w^2 + 3w = 90
Hopefully, that part makes sense to everyone as far as it is a formula for the area of the deck.
... before we can use the ZPP, we need to set the equation = zero, so...
which can be simplified to
2w^2 + 3w - 90 = 0 (still no magic, right?)
This CAN be factored OR we can use the Quadratic formula
If we factor, we end up with
(2w + 15)(x - 6) = 0
which presents us with the solutions:
w = -7.5, 6
Since the context of the problem DEMANDS a positive solution, the only USEABLE solution is 6, since a rectangle can't be measured with a negative length, therefore the solution of -7.5 is EXTRANEOUS (i.e. irrelevant).
So, w=6 means that the width equals 6, and we go back to the original problem to see that the length is 2w+3, or in this case 15.
SO, the solutions to the quadratic mathematical model are -7.5 & 6, of which 6 is the only one that makes sense in the context of the real life problem.
SO, the ANSWER to the question is that the width is 6 and the length is 15.
It might seem "nit-picky" but do you see how I am using the words SOLUTION and ANSWER in different ways.
We SOLVE math (model) problems... we ANSWER real-life questions, and we sometimes have to pick through the "weeds" of the SOLUTIONS to the mathematical model to find our ANSWERS.
What FUN???!!
Mr.C I'm, having trouble on #11 on the Chapter test. I said the length was w+15. My quad formula was w(w+15)-100=0. After I put it in the quad equation I got w=20 and the length would equal 35 which DOES NOT equal 100. I think the right answer is w=5 and L=20 but I don't know how to get that!! Oh and btw #14 makes sense now. Thanx!
ReplyDeleteOH MY GOSH!!! I made a really big mistake in setting up my quad formula and forgot to make the b negative, so the 15 was postive!! Opps, now it make sense!!!!
ReplyDeleteThat's actually a pretty easy mistake to make... your painful evening of study will undoubtedly be rewarded... good luck tomorrow!
DeleteWho knew... that MATH could be THIS exciting!!
ReplyDeleteMr.C I don't understand how to solve #21 on pg 617 on hw #10-1. I tried mapping the joggers path into a triagnle and plugging in the numbers for the Pythagorean Theorem, but it didn't come out to the right answer. Could you help???
ReplyDeletepls see the latest blog post for 10-1 & 10-2, i moved your question there...
Delete