Thank you.
John Whitaker
Type: Posts; User: John Whitaker
Thank you.
John Whitaker
I''m sorry... I can do many problems in complex numbers, but I have not seen one structured like this one. I learn not by reading complicatiions, but by following similar examples step by step.
...
Only instruction is to "Solve these equations>"
"Hornsby, Lial, & McGinnis" 8th Ed.
My instructor likes to create problems for which there are no examples or similar problems in our text we might refer to for clarity. So I need to learn how to deal with:
4x^2 + 1 = 0
4x^2 ...
I think I get it. Thank you.
I have:
2x^2 + 12x + 3 = 0 (divide by 2)
x^2 + 6x + 3/2 = 0
x^2 + 6x = - 3/2
X^2 + 6x + 9 = 15/2
(x + 3 )^2 = 15/2
x + 3 = +/- sqrt (15/2)
x + 3 ...
Thank you.
I get just so far before becoming confused.
2 [ 5 - ( - 3 ) ] - 2 ( - 3 - 4 )
2 [ 5 + 3 ) - 2 ( - 7 )
2 ( 8 ) - 2 ???
Minus 2 x minus 7 (or is it a plus 7 ?)
I can't be certain, and I have...
I'll try to figure it out, though I don't recall anything in my text that relates to the assignment or that exponent. I'll look again.
Thanks
John Whitaker
I don’t (yet) know how to illustrate the radical has an index, so I wrote it in. Perhaps I am about to learn. However,
index=4 sqrt 5x / z = index=4 sqrt 5x / index=4 sqrt z * index=4 sqrt z^3...
Gentlemen,
I do appreciate your efforts... and I'll study what you have shown me, though it still is not as clear as I require (not your fault). I won't post to this again. Thank you.
John Whitaker
Somewhere my problem has been turned around and solutions are being given to the part I understand. I’ll begin again. Please be patient. I can’t move ahead until I understand this
K^2 + 5K -1 ...
Denis, Thanks, I'll try it. Don't give up.
Mark, Thanks. k(k+5/2) = k^2 + 25/4 yes... no...?
WAIT! I think I'm getting it...! I have been forgetting to add the k to the 5/2 part of the...
I thank you for your input, but I'm not smart enough to grasp it. I worked it out substituting the numbers and it did not clarify my problem.
I will appreciate it if, using the numbers I...
I' m sorry... I thought I could figure it out from the responses to my first submission, but I can't. The hang up is that k^2 + 5k + 25/4 = 29/4.
The next step is (k + 5/2)^2. I don't see...
I' m sorry... I thought I could figure it out from the responses, but I can't.
(k + 5/2)^2 returns k^2 + 25/4... not k^2 + 5k + 25/4 that I had going in.
Where is the middle term? I'm...
Denis,
I'm still struggling... hope this is the last semester. Thanks. (You haven't qiven up either).
John
Gentlemen,
Thank you and I wish you a great holiday.
John
The problem:
Solve k^2 + 5k -1 = 0
k^2 + 5k = 1
k^2 + 5k + 25/4 = 4/4 + 25/4
k^2 + 5k + 25/4 = 29/4
The next step my text displays is (k+5/2)^2 = 29/4 without telling me how it arrived at...
I think I have it. Thank yoy.
John Whitaker
I'm sorry... still don't get it. Can you show me where I am to do this?
Thank you.
John Whitaker :wink:
The problem and book solution is:
2 sqrt 75/16 + 4 sqrt 8/sqrt32
= 2 sqrt 25*3/sqrt 16 + 4 sqrt 4*2/sqrt 16*2
= 2(5 sqrt 3/4) + 4(2 sqrt 2/4 sqrt 2)
= 5 sqrt 3/2 + 2
= 5 sqrt 3/2 ...
Thank you, Denis...
Another: My text, Lial, Hornsby, et al, assumes I know more than I do and takes leaps over steps I need. I get down to:
X^-8/3-4/3 Y^4-(-2/9)
and I don't understand how...
Loren,
Makes sense. Thank you.
John Whitaker
It's been a while, and I forgot how to express a radical.
Instruction: Write each radical as an exponential and symplify.
Problem displays Y^10 under a radical. Answer given is Y^5. What are the...
Thank you.
Signing out.
John W