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?

1. a^2 - a = 12
2. b^2 - b = 2

Please explain your answers.

(Edit: Added the original question for clarity.)

mst · Answer

I get C.

From 1, a = 4 or -3
From 2, b = 2 or -1.

sqrt ( a^3 - a^2 - b) = 7 is satisfied only by a = 4 and b = -1

Fig · Answer

for me (C)

sqrt ( a^3 - a^2 - b) = 7 => a^3 - a^2 - b = 49 (1) Stat 1 : a^2 - a = 12 <=> a^2 - a - 12 = 0 <=> (a+3)(a-4) = 0 <=> a = -3 or a = 4 Both values of a work in (1) INSUFF Stat 2 : b^2 - b = 2 <=> b^2 - b - 2 = 0 <=> (b-2)*(b+1) = 0 <=> b=2 or b=-1 Placed in (1), we have several possible values of a INSUFF Combining (1) with (2): case 1 : a = -3 (1) => (-3)^3 - (-3)^2 -b = 49 <=> -36 - b = 49 <=> b = -85 >>> Not matching values of b in statment 2 case 2 : a = 4 (1) => (4)^3 - (4)^2 -b = 49 <=> 64 - 16 - b = 49 <=> b = 48-49 = -1 >>> Bingo

1 match with 1 value of b in statment 2

muddypaws · Answer

B
as mst and fig have calculated
from stat 2, b = 2 or b=-1

from stem a^3-a^2-b=49
hence a^3-a^2 = 49+b
a*a*(a-1) = 49+b

if b = 2
a*a*(a-1) = 51
but 51 = 17*3 and neither of them can be used as a in equation above (and we have to find integer a)

if b=-1
a*a*(a-1) = 48

factoring 48, it is possible to get
48 = 4*4*3
hence a=4

from stat 1, a=4 or a=-3
both can be used to satisfy the stem equation and arrive at different solutions for b as b = a*a*(a-1) -49 where b= -1 or b=-85

ugo_castelo · Answer

I'm getting B as well
the only option for a and b to be integers is if a=4 and b=-1

Fig · Answer

Yesss

..... It's (B)

This is my favorite silly mistake

: to forget that we are speaking about integers. As a is in integer, there is only 1 possible solution with the statment 2

Thanks guys :D

anandsebastin · Answer

Muddypaws nailed it. 

OA is B. The explanation was scrambled in Princeton Review.

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

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