Ok, so we're just going to complete the square... no bunnies... no eggs.
Please confirm your answers using a graphing calculator (TI-84 or Graphit on the web).
The problems that have a 'b' coefficient that is odd (i.e. #22: p^2+5p-7=0) are a little tougher, but I expect you to PERSIST and find a solution (or ask me a question). The method is the same, you just have slightly more difficult numbers to work with... you have a calculator... stop whining!!
pg 564-66 #5-27 Odd, 33-39 Odd, 46,47
I AM SO CONFUSED ON EVERYTHING!!! I DON'T KNOW WHAT TO DO!!! PLEASE HELP!!!
ReplyDeleteHello Ryan... so nice of you to check-in... you missed a key lesson "Solving Quadratics by Factoring and the Zero Product Property" on Monday, so it's not unusual that the the lesson on "Solving Quadratics by Completing the Square" on Weds seems that much more intimidating.
ReplyDeleteI'm sure there might be a few of your alge-buddies feeling the same pain... so, let's talk REMEDIATION.
I can be available before school on Mon & Tues of this week... let me know if you (ANYONE) want to stop in around 7:15 or so... I highly recommend a visit!
What are you (EVERYONE) doing to self-remediate?? Have you watched the Pearson Video Tutors on MathChamber?? I recently added Khan Academy videos on MathChamber... check those out... I would like your (EVERYONE'S) opinion on the Khan vs. Pearson videos.
All that said, can you give me a more specific question than the SHOUTING YOU OFFERED ABOVE??
Mr. C.
Would it be O.K. to use the tile link on mthchamber to solve a few of the problem? (exe. 7, 9, 11, )
ReplyDeleteABSOLUTELY!!!!! I'm so happy that you see the connection!!!!!!!!!!!!!!!!
ReplyDeleteI'm having trouble with the 13-27 area. I'm using the process described in 9-5, but it never comes out as anything. What am I doing wrong?
ReplyDeleteI want you to watch the video tutors, Pearson & Khan Academy, and let me know it they help.
ReplyDeleteProblems with even (as opposed to odd) 'b' coefficients tend to be a little easier, but ALL can be solved using COMPLETE THE SQUARE.
#15)
m^2+16m=-59
ok, so if we want to turn the left side into a perfect square trinomial, we need a 'c' value, right.
If we used the area model, we would have m+8 on both sidebars in order to sum to 16m, right? If we square (m+8), we will get 64 for 'c', right?
So,
m^2+16m+64=-59+64, equivalent, yes?
So,
m^2+16m+64=-59+64
(m+8)^2=5
Take the sqrt of both sides
m+8=+/- sqrt(5)
m=-8 +/- sqrt(5)
m={-10.24, -5.76}
Graph the original equation using a graphing calculator... do the roots make sense?
lmk (and watch the videos)
#13) a little tougher cuz of the odd 'b'
ReplyDeleteg^2+7g=144
ok, so if we want to turn the left side into a perfect square trinomial, we need a 'c' value, right.
If we used the area model, we would have g+(7/2) on both sidebars in order to sum to 16m, right? If we square (g+7/2), we will get 49/4 for 'c', right?
So,
g^2+7g+(49/4)=144+(49/4), equivalent, yes?
So,
g^2+7g+(49/4)=(576/4)+(49/4)
[g+(7/2)]^2=625/4
Take the sqrt of both sides
g+(7/2=+/-(25/2)
g=-(7/2) +/- (25/2)
g={-9, 16}
Graph it and see...
Graph the original equation using a graphing calculator... do the roots make sense?
My questions will come tomorrow.. I've been so busy!
ReplyDeleteThis is what I got for # 27
ReplyDelete5n^2-3n-15=10
5n^2-3n+9/4=27 1/4
----------------------
5
n^2-12/20n+9/20=5 9/20
I don't think I can solve this, maybe I am doing things in the wrong order.
Also, I need some help with 33
ReplyDeleteFor #5, I got the opposite answers than in the book, a & b were switched.
ReplyDeleteFro #27, adding 15 to both sides was a reasonable first step, but you should have divided both sides by 5 as the next step. THEN take half of the middle coefficient and square it.
ReplyDelete#33 is a toughie, we'll leave it for class... the trick is in defining the variables and creating the expressions.
I don't understand the comment re: #5.
For 5a:
k^2-3k=304
k^2-3k-304=0
(k-19)(k+16)=0
So, that was just the factoring step!
The values of k that make the equation true are (aka THE SOLUTIONS!!!):
k={19,-16}
Was that your mistake?
5b should have been done via Complete the Square
t^2-6t+16=0
t^2-6t+9=-7
(t-3)^2=-7
... and you would find NO REAL NUMBER SOLUTION, yes? (since there is no REAL NUMBER that is the square root of -7)
For #13, would I square 3.5 to get the 3rd term?
ReplyDelete@Anonymous with really long answer-
ReplyDeleteAre you sure you're not Mr.Chamberlain? =D
If the answer ends up being something like 2.455677, does that mean there is No Real Solution?
ReplyDeleteI'm having trouble with #15.. This is hard! D=
ReplyDeleteNone of the numbers in, for example, (x+9)^2=388, can be squared to a real number! Like 388!
ReplyDeleteThe question above is for #17..
ReplyDeleteFinally! One that is Square Root-able! #19!
ReplyDeleteBy the way, are we going to have NJASK review?
ReplyDeleteI don't understand #23..
ReplyDeleteI keep getting stuck on problems like (a+5)^2=30! What do we do if the number after '=' can't be squared, like 30?
ReplyDeleteWhat do I do for #25, where A=4?
ReplyDeleteOops, sorry, I didn't see the section that said "Completing the Square when A does not equal 1" =D
ReplyDeleteFor #27, when I'm finding half of B, and B is -(3/5), what do I use for B to get the 3rd term? 1.5/5?
ReplyDeleteI would like to come in for extra help..
ReplyDeleteAnd also, on #33, I don't understand where to go, the equation I have for the width is w=600/l
I watched a few Khan Academy videos, but I like Pearson better because it's not as long & it's more "To the point".. Also they have neater handwriting =)
ReplyDeleteI don't understand the directions for #46, 47.
ReplyDeleteI will come in early on monday for extra help
ReplyDeleteI'm so confused on #33! I got 2 equations for w, (75-L)/2=w and 600/L=w. I don't know how to solve them. When I tried to substitute like in a system of equations I couldn't get an answer one way and I got a negative number the other way...??? can we go over this in class
ReplyDeleteI agree with Julia...
ReplyDeleteAnd to all my questions, too...
We need answers, Mr.C! Answers!
And I am most likely coming in for help tomorrow morning!
Based on your feedback, we will clearly be reviewing several items in class.
ReplyDeleteI think many of you are confusing a perfect square with taking the square root of a number.
The square roots of 1,4,9,16,25,36,... are 1,2,3,4,5,6,etc. , so these numbers (1,4,9, etc.) are considered PERFECT SQUARES, since their square root is an INTEGER.
Numbers like 2,3,5,6,7,8,10... can be "square rooted" but they are NOT perfect squares. It turns out that the square roots of these numbers are irrational numbers. Irrational numbers ARE A SUBSET of the set of REAL NUMBERS. We will discuss this fact in more depth in our next unit.
So, for a problem like:
(a+5)^2=30
you take the sqrt of both sides and get
a+5=+/- sqrt(30)
a=-5 +/- 5.48
a= -10,48, 0.48
I will be in tomorrow morning by 7am or so... would anyone like to join an alge-chat now??
ReplyDeleteI would like to join alge-chat!
ReplyDeleteAnd I am coming tomorrow!
If you go to the MathChamber Algebra Table of Contents page, you can join us in a chat by clicking on the red VIRTUAL CHAT link...
ReplyDeleteI won't be there for too long... it's already past my bedtime!
Mr.C? Are you still on chat?
ReplyDeleteI'll see you guys in the morning... and at lunch!
ReplyDelete