aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given in the question stem that \((a^3-a^2-b) = 49\)\(\to a^2(a-1) = 49+b\)
From F.S 1, we know that a(a-1) = 12, thus, \(a*12 = 49+b \to\) a is an integer for both b = -1 or b = -13. Thus, we get two different values of a, Insufficient.
From F. S 2, we know upon solving for the quadratic \(b^2-b-2\), the integral roots are -1 or 2.
Now,\(a^2(a-1) = 49+b\)
For b = 2, \(a^2(a-1) = 51\). Assuming that a is odd/even, the given product is an arrangement like odd*odd*even = even, and 51 is not even.
Similarly, for a is even, the arrangement will be like even*even*odd = even, and just as above 51 is not even. Thus, \(b\neq{2}\)
For b = -1,\(a^2(a-1)\) = 48. Now all the integral roots of this polynomial can only be factors of 48, including both negative and positive factors.
However, any negative factor will never satisfy the given polynomial as because (a-1) will become a negative expression, which can never equal 48.Hence, the given polynomial has no integral solutions in -48,-24,-12,-8,-6,-4,-3,-2,-1.
By the same logic, we know that any integral solution,if present will be one of the positive factors of 48.It fails for 1,2,3 and we find that a=4 is a root.If for a=4, the expression equals 48, then for a value above 4, the expression \(a^2(a-1)\)is bound to be greater than 48.
Thus, the only solution possible for the given polynomial is a=4, a unique value,Sufficient.
B.
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