For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the

can you elaborate on how you solved \(a^2(a-1) = 48\) to \(a = 4\)?

Similarly how did you conclude that \(a^2(a-1) = 51\) will not have an interger as a solution?

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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.

Tough one...Took more than 3 min 50 sec and ended up getting it wrong...

We can change the given equation to (a^3-a^2-b) = 49 (squaring both sides)

From St 1 we have a^2-a=12 ----> substituting in above given eqn we get
{a(a^2-a)- b} =49 ------> 12a-b=49 -----> a = (49+b)/12
Now a and b are integers therefore (49+b)/12 should be an integers

Possible values b=11, a=5
b=23, a =6, b=-1, a=4
So St 1 alone is not sufficient

St 2 says b^2-b =2 -----> b(b-1) =2 possible values of b are

b=2 or b=-1
Substituting in given expression we get
(a^3-a^2-b) = 49

a^3-a^2= 51 or------> a^2(a-1)= 51 (17*3) we see that 51 even after reducing to prime factors gives us no Integer value of a

a^3-a^2= 48 -----> a^2(a-1)= 48 (4^2)*3 and hence we get value of a= 4 or -4

Substituting values of b=-1 and a=4 or -4, we can see that only for a=4 the above given equation is proven.

Hence a =4

I did highlight the value of b=-1, a=4 from statement and can be taken as a hint without solving completely eqn 2
Ans B

Minor mistake. The given polynomial will not yield a = -4 as a root.

if \(b=-1\) then \(a^3-a^2=48\) --> \(a^2(a-1)=48\) --> \(a=4=integer\)

how do you solve something like that, is it just quick trial & error since you know a & b have to be integers and there's only a few values of a that would give results somewhat near 48?

Yes, that's correct. We know that \(a\) is an integer, thus \(a^2(a-1)=(perfect \ square)*(positive \ integer)=48\). From here you can use trial and error and find that \(a=4\).

I have just one question, why isnt the absolute value used after the expression is squared?

\(\sqrt{x}=2\) --> \(x = 4\). Where should there be an absolute value?

(1) a^2 - a = 12

a can be 4 or -3

Either case can make the provided equation hold since b can be any number.

(2) b^2 - b = 2

b can be either 2 or -1

If b = 2, then a^3 - a^2 = 51
If b = -1, then a^3 - a^2 = 48

The first equation is not possible since the result is an odd number. a^3 - a^2 must yield an even number either if a is odd or a is even.

The second equation we get a = 4.

SUFFICIENT.

The answer is B.

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