Results 1 to 14 of 14
Thread: New and clever SAT problem

20180415, 23:14 #1
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,701
 Thanks
 60
 Thanked 89 Times in 82 Posts
New and clever SAT problem
For your solving minds...

20180417, 12:19 #2
 Join Date
 Feb 2008
 Location
 A cultural area in SW England
 Posts
 3,584
 Thanks
 35
 Thanked 218 Times in 193 Posts
Nobody seems to have bitten yet, so I say
BATcher
Existence is useful.

20180417, 12:24 #3
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,701
 Thanks
 60
 Thanked 89 Times in 82 Posts
Well, (a) that's not an answer option and (b) it's not the case. There is a specific value for b.

20180424, 06:46 #4
 Join Date
 Dec 2009
 Location
 Surrey, UK
 Posts
 285
 Thanks
 13
 Thanked 70 Times in 60 Posts
I am puzzled: I can work out that setting b to one of those values makes the expression simple (and I know which one), but cannot see why it has to be so. It all boils down to the exact meaning of the phrase "is equivalent to bx". If that means simply numerically equivalent, then (surely?) the equation rearranges to:
x^2(4a  1) + x(4a4b) = 0
and the solution for x is that it equals 0 or (4a4b)/(4a1). So 'b' can have any value we like. But that is clearly not what the puzzler intends. If 'equivalent' means 'algebraically equivalent' (if that is a valid term), then
x^2(4a1) + x(4a4) == bx
and then
Maybe I need to go and read some SAT test rules, but I don't like this! I may have to accept I am just not clever enough. I am certainly not clever (or informed) enough to create a spoiler box as BATcher did  oh, I just found out something  who knew that option was there? (I did a reply with quote to see what Batcher had used).
Martin
PS Glad to see a puzzle here after so long. I do have one of my own I rather like...Last edited by mngerhold; 20180424 at 06:48.

20180424, 11:01 #5we know a must be 1/4
I used to be good at this sort of thing
The stein is BACK
(4x+4)(ax1)x^2+4=bxLast edited by wavy; 20180424 at 11:12.
David
Just because you don't know where you are going doesn't mean any road will get you there.

20180424, 11:17 #6
http://www.wolframalpha.com/input/?i...t+does+b+equal
wolframalpha could not get it eitherDavid
Just because you don't know where you are going doesn't mean any road will get you there.

20180424, 13:05 #7
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,701
 Thanks
 60
 Thanked 89 Times in 82 Posts
This is, what I consider, an awful problem from the College Board. However, the "key" to solving this is realizing that "b" is a constant (one of the values in the answer choices) and is the coefficient of "x". Therefore, you have to clear everything on the left side of the equation EXCEPT for some number of "x" values.
That means:

20180424, 13:17 #8
 Join Date
 Feb 2008
 Location
 A cultural area in SW England
 Posts
 3,584
 Thanks
 35
 Thanked 218 Times in 193 Posts
I'm not sure that you have told us what is the value of b, which is the original question!
BATcher
Existence is useful.

20180424, 13:35 #9
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,701
 Thanks
 60
 Thanked 89 Times in 82 Posts
Last edited by kweaver; 20180424 at 13:38.

20180531, 13:16 #10
x has to be 1, in order for b to be 3. If these are true, then you end up with 4a + 1 = 1  4a.

20180531, 17:08 #11
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,701
 Thanks
 60
 Thanked 89 Times in 82 Posts
Since the problem stated that the expression equals "bx" you're then looking for the coefficient of the "x" term. When you expand the problem, the 4 and +4 cancel. That, I suspect, was the College Board's attempt at a clue. The x^2 terms must also be eliminated because the expression = "bx" and there are no x^2 terms.
So, if 4ax^2  x^2 must = 0 then, 4ax^2 = x^2 and 4a = 1, so a = 1/4. Consequently, 4ax  4x becomes 4(1/4)x  4x = x  4x = 3x; therefore, b = 3.

20180601, 09:09 #12

20180603, 16:47 #13
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,701
 Thanks
 60
 Thanked 89 Times in 82 Posts
Don't think so. Since the expression was equal to "bx", you need to convert the expression to have only an "x" variable. That means eliminating the constants and the x^2 terms.

20180604, 09:21 #14
Of course, you can't just eliminate the constants and the x^2 terms; you have to eliminate them in a mathematicallyvalid way. I'm sure you know that; I just thought I would point it out.