A Radical Prompt
Blogging Prompt
Here's the Blogging Prompt we discussed in class today:
Without using a calculator find the value of:
sq.rt.(32 + 42 + 122) sq.rt.(7 + sq.rt.(4))
Your challenge, should you choose to accept it, is to create two radical problems of your own similar to those above. This challenge is open to all takers! However, once a solution is posted you CANNOT use the same solution as your own -- you've got to come up with a new one. In other words, the longer you wait to tackle this, the harder it's going to get.
Write your answers in the comments to this post. You can also use cube roots (cube rt.), fourth roots (4th.rt.) and higher if you wish. Feel free to mix them up too! (i.e. square roots and cube roots, or square roots and fourth roots, etc.) A successful reply to this prompt includes TWO answers; one similar to each of the problems posted. If you have a successful reply you do not need to write a reflective pre-test blog post.
Have Fun!
posted by Darren Kuropatwa @ 11/07/2005 02:34:00 PM
17 comments
17 Comments:
sq. rt.(4^2+9^2+7^2)
sq rt.(6+sq. rt.(14))
for the second one is
sq. rt.(6+sq. rt.(14))
sorry forget the dot
Sorry Jan, try again.
The first one equals sq.rt.(146), which IS NOT a whole number.
In the second one sq.rt.(14) IS NOT a whole number.
a) sq. rt. (5² + 12²0
= 25 + 144
= 169
sq. rt. = 13.
b) sq. rt. (13 + sq. rt. 9)
= 13 + 3
= 16
sq. rt. = 4
sq. rt.(20^2+12^2+16^2)
sq. rt.(6+ sq(9))
So far NO ONE has earned that blog prompt mark.
Thang - you heard me in class, right?
In order to get that mark you need to have TWO problems, one modeled after each of the ones I presented, that can be solved as whole numbers. SOme of you have one correct but not the other.
If you have already posted your pre-test reflective blog you already have your blogging mark. (i.e. Janet)
sq. rt.(12^2+4^2+3^2)
sq. rt.(6+sq. rt.(9))
1.) SQ.RT.(2^2 + 6^2 + 9^2)
= SQ.RT.(4 + 36 + 81)
= SQ.RT.(121)
= 11
2.) SQ.RT.(4 + Sq.Rt.(144))
= SQ.RT.(4 + 12)
= SQ.RT. (16)
= 4
PLEASE BE RIGHT!!!! QUICK QUICK QUICK RACE!
SAMUS you got it but your solution to the first one is incorrect, please repost and fix it.
Jojo you also have both right but you second problem was presented earlier by someone else. Repost. Keep your first one; create a new second radical.
EVERYONE READ THIS:
This: sq. rt.(12^2 + 4^2 + 3^2)
and this: sq. rt.(12^2 + 3^2 + 4^2)
and this: sq. rt.(4^2 + 12^2 + 3^2)
etc.
ARE ALL THE SAME AS THE ORIGINAL RADICAL I GAVE YOU.
tim-math-y, yours is perfect! You got the mark.
Again, EVERYONE, please take note:
When you repost your problems repost them BOTH -- don't make me hunt through the earlier posts to verify your work.
You guys are all doing some really good thinking! Don't give up; you CAN do this!!!!!
lol this is hilarious xD
b) sq. rt. (36+ sq. rt. 9)
= 6 + 3
= 9
sq. rt. = 3
Mr K, hope you are happy with my questions.
The people who have got their blog mark for correctly replying to this prompt are:
tim-math-y
SAMUS
The rest of you either have both or one of your problems incorrect. The errors are either;
(a) mathematically incorrect or
(b) you didn't create both problems SIMILAR to the ones I posted.
Don't give up! Keep trying. You CAN do this!! ;-)
b)
sq. rt. (36+ sq. rt. 9)
= 6 + 3
= 9
And the sq. rt. of 9 is three.
Maybe that's it. I hope so.
PS. Everyone click on my name, lol.
a)
sq. rt. (5² + 12² )
= 25 + 144
= 169
sq. rt. = 13.
Maybe that's it. I forget the other bracket.
or
sq. rt. (5² + 12² + 3³)
= 25 + 144 + 27
= 196
sq. rt. = 14.
Hee hee... By jolly, I think I got it.
Not yet Thang -- use either all squares or all cubes. You can try to find a cube root of a sum of three squares if you like, or something similar to that.
Your other problem and solution is incorrect.
Please go back and READ ALL MY EARLIER COMMENTS on this post.
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