DS: Value of a

a^2-a=12--->a^2-a-12=0 ---> (a-4)(a+3)= 0 ---> a=-3,4
b^2-b=2---> b^2-b-2= 0 ---> (b-2)(b+1)=0 ---> b=-1,2
now look at the square root, quickly you can eliminate a=-3 , becoz a^3 is the biggest power and the determinant of the value of the squareroot, if a=-3, the value inside the sqrt<0 ---> undefined.
Test a=4, try b=-1 , yeah, that's it!
try b=2, no.

----> a=4, b=-1

Do you guys say the answer is C? The OA is B, but I don't know how to conclude that.

okie, let's see:
from 2, we obtain b=4 or -1
that is to say : the square root can be 7^2+(-1) (= 48) or 7^2+ 2(=51)

case 1:
we have a^3-a^2 = 48
---> a^3 + 3a^2- 4a^2 + 12a -12a -48= 0 ( this is a high school's method to factorize the expression to obtain an equation with RHS= 0)
---> a^2(a-4) + 3a(a-4) +12(a-4)= 0
----> (a-4) * ( a^2+3a+12)=0
-----> a=4, a^2 + 3a +12 has no solution since delta<0 ( quadratic equation method)

case 2 : a^3- a^2= 51 or a^3-a^2-51=0
since a must be an integer ---> the possible roots of this equation must be among factors of 51 ( +-17 and +- 3) . You can plug these numbers to see if they're roots of the equation . NOTE: Better method to plug is to right away notice that the root can't be negative because the highest power of a here, a^3 determine the value of LHS. After plugging, you find that there's no root for this equation.

Thus, the only value of a obtained is 4 when b=-1

Thus, B is correct!

this is a good question ! what is the source of it ?

very good question. almost got this one wrong.

got blown away on this one...

now I see why it is B...lexi nice job!