For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is

Question

For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is the value of a?
I) a^2 - a = 12
II) b^2 - b = 2

venksune · Answer

1.a=4 or-3. Two values for a and hence insuff.
2. b can be 2 or -1. Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4. 

hence B.

sofere · Answer

Please explain your method in solving this. My calc with polynomial equations is hazy. Thanks.

twixt · Answer

I think it is 49 = a.(a^2-a)-b so a = (49+b)/12

knowing that (stat. 2) b = 2 or -1, a being an integer, b = -1 and a = 4

saurya_s · Answer

I think C
1 gives 12-b=49
2 gives quadratic b=-1 and 2
Only b=-1, gives a as integer=4
S

twixt · Answer

I think you just prove statement 2 is suff !

saurya_s · Answer

how do you solve
a^3-a^2-48=0 to find the value of a to conclude B as answer?
S

sdanquah · Answer

Hi Venk, I see that your solution for a in statement 1 neglects the original statement which states that if sqrt(a^3 - a^2 - b) = 7, what is the value of a, You solve statement 1 independ of the original equation and that does not seem right. 
At best, from the original equation we can say that 
a^3 - a^2 - b = 49
Now given 1, a2-a=12 implies that  12a -b =49  from (a(a^2-a) -b =49, making statement 1 insufficient.
Statement 2 now gives you two values of b (2, -1) which when substitude into the original equation yield 2 values, one an integer of 4 and the other a non integer. Thus for a to be an integer it has to be 4.  B is sufficient

sdanquah · Answer

It is posible, try picking numbers adnd you will land at a = 4 or a = 3.25.
It is easy to fall into the trap that 
statement 1 combine with the original statement yields
12a -b = 49 and thus the moment one gets b, one can find a making C the dangerous answer. But if you substitue the value of B calculated and substitude into the original equation you might get only one integer for a and that is 4.

venksune · Answer

sdanquah,
I am also saying only B is sufficient. In stem 1, we have two values of 'a' - question is what is the value of 'a'. We cant conclusively say the value of 'a'. Also, when applied to the original equation, we still need 'b'. I didnt neglect it, I thought the added data is not necessary to prove insufficiency of stem 1.

amernassar · Answer

for the second we have
b=-1,2

for b=-1 , sqrt(a^3-a^2+1)=7. implies a^3-a^2+1=49

or a^3 - a^2 -48 = 0
a = 4 is a solution
divide a^3-a^2-48 by a-4 gives the polynomial a^2+3a+12

so (a^2+3a+12)(a-4)=0
a^2+3a+12 is never zero since discriminant is negative (no roots for a) :-D

giving a =4

for b = 2

we get a^3-a^2=51
a^2(a-1)=51

from this we cannot find any value of A

hope this helps.

giving a =4 for b = 2 we get a^3-a^2=51 a^2(a-1)=51 from this we cannot find any value of A hope this helps.

hardworker_indian · Answer

OA is B. From Princeton.
OE = "Statement 2 gives b = 2, -1. Since a cant be an interger if b = 2, it must be true that b = -1 and a = 4."

sdanquah · Answer

Agreed with your answer venksume, Just that I found that you neglected orignial statement in your earlier solution.

For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is

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