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by Gurpinder » Tue Aug 31, 2010 9:51 am
If #p# = ap3+ bp - 1 where a and b are constants, and #-7# = 3, what is the value of #7#?

A)5
B)0
C)-2
D)-3
E)-5


I see that #-7# = 3 means:
#-7# = A(-7)3 + B(-7) - 1

and so #7# means:

#7# = A(7)3 + B(7) - 1

But now what.....
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by this_time_i_will » Tue Aug 31, 2010 10:03 am
expanding as per the given formulae:
#-7# = 7^3a+7b+4=0
Now, #7# = 7^3a+7b-1 = 7^3a+7b+4-5 = 0-5 = -5
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by Gurpinder » Tue Aug 31, 2010 10:06 am
this_time_i_will wrote:expanding as per the given formulae:
#-7# = 7^3a+7b+4=0
Now, #7# = 7^3a+7b-1 = 7^3a+7b+4-5 = 0-5 = -5
Um.....what formula?

#-7# = 7^3a+7b+4=0

How'd you do all of that?
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by Makushr1 » Tue Aug 31, 2010 11:38 am
Gurpinder wrote:If #p# = ap3+ bp - 1 where a and b are constants, and #-7# = 3, what is the value of #7#?

A)5
B)0
C)-2
D)-3
E)-5


I see that #-7# = 3 means:
#-7# = A(-7)3 + B(-7) - 1

and so #7# means:

#7# = A(7)3 + B(7) - 1

But now what.....
Here's what I did:

a(-7)^3 - 7b - 1 = 3
-334a - 7b = 4
-7(49a+b)=4
49a+b=-4/7

Plug in 7 now

a(7)^3 + 7b - 1 = ?
334a+7b - 1 = ?
7(49a+b) - 1 = ?
Since we know from the first part that 49a+b=-4/7, plug that in.
7(-4/7) - 1 = ?
-4-1=-5
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by Arcane66 » Tue Aug 31, 2010 12:24 pm
Here is how I solved it:

If #p# = ap3+ bp - 1 where a and b are constants, and #-7# = 3, what is the value of #7#?

First step is to plug in -7 as p into the equation given above. I got 3 = -21a +-7b -1. From this, add 1 to both sides and then divide everything by -7 to get the equation -(4/7) = 3a+b. All I did from here is to simply guess what a correct combination would look like. I plugged in zero for a and -(4/7) for b. Then, I used the value of 7 for the original equation and used my constants that I came up with to get an answer of -5.

The answer, in my opinion, is E.
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by OliverTPR » Tue Aug 31, 2010 10:48 pm
Gurpinder,

You are on the right track. The catch to this problem is to realize that you cannot, nor need to, solve for the constants. The other catch is to notice that we are trying to solve for the negative input of an input with a solution. (7 instead of -7.)

First you put in the -7 for the input p and solve to get the equation you mentioned.

#-7# = A(-7)3 + B(-7) - 1

Now, we also know that #-7# = 3, per the problem. Let's put that in. (I am also going to calculate everything out, so it is nice an pretty looking.)

3 = -21A - 7B - 1

Now, we should do the reasonable next step and move that -1 over to the other side to isolate our variables.

4 = -21A - 7B

Okay, now let's put the 7 into our function and see what the problem looks like.

#7# = A(7)3 + B(7) - 1

Calculate it all out and you get...

#7# = 21A + 7B - 1

Notice that the 21A + 7B kinda looks familiar? This is because know the solution for -21A - 7B from earlier. Since we have some information that is a little different, we need to use some strategy and make it look the same. How does -21A - 7B change into 21A + 7B? Divide by -1. So...

4/ (-1) = (-21A - 7B)/(-1) =
-4 = 21A + 7B

So, now we can plug and solve our new equation.

#7# = 21A + 7B - 1

becomes...

#7# = (-4) - 1 = -5


I hope this helps you out. It is a little longwinded, but I wanted to make sure that you followed the logic. Let me know if you have any questions!
Oliver Pope
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The Princeton Review
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by Gurpinder » Wed Sep 01, 2010 6:51 am
Thank you Oilver!

That clarifies all doubts! :D
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