For integers a and b if square root of (a^3-a^2-b)=7,

I am getting D).

Whats the official answer?

(1)
a^2-a-12=0
(a-4)(a+3)=0
a= 4 | a= -3

We can't determine from the information given whether a will be -4 or 3 because depending on what b is either of these values are possible.

(2)
b^2-b-2=0
(b-2)(b+1)=0
b= 2 | b=-1

If b is 2 then no integer for a will make the sqrt come out to 7.

therefore b must be -1 and knowing b is -1 the only possible integer for a that will make the sqrt=7 is 4

I pick B

a^2-a-12=0
(a-4)(a+3)=0
a= -4 | a= 3

Dmateer,

(a-4)(a+3)=0

a= -4 | a= 3 should this not be a=4 and a=-3

Only 4 works for a^2-a = 12

If a=3 3^2-3 = 6 and not 12

Let me know ur thoughts

Hence I choose D)

Actually my approach was

a^2-a = 12
a(a-1) = 12

only possible when a=4 a-1=3 since its given a and b are integers

For integers a and b if square root of (a^3-a^2-b)=7,


What is the value of a?

(1) a^2-a=12
(2) b^2-b=2

In DS questions, both statement always give us TRUE facts, either we need them or not!

Statement 1 tells us: a=4 or a=-3

Statement 2 tells us: b=2 or b=-1

So actually we have 4 possible pairs for (a,b)

(4,2) (4,-1) (-3,2) and (-3,-1)

Let's put these values in our equation:

square root of (a^3-a^2-b)=7

Only (4,-1) gives us the solution.

This solution can be obtained from each statements. So do NOT choose (D)

Read the posts below:

cramya wrote:

a^2-a-12=0
(a-4)(a+3)=0
a= -4 | a= 3

Dmateer,

(a-4)(a+3)=0

a= -4 | a= 3 should this not be a=4 and a=-3

Only 4 works for a^2-a = 12

If a=3 3^2-3 = 6 and not 12

Let me know ur thoughts

Hence I choose D)

Actually my approach was

a^2-a = 12
a(a-1) = 12

only possible when a=4 a-1=3 since its given a and b are integers

Cramya, yes you were correct a = 4 and a = -3. I did this correct on my scratch paper but typed it incorrect.


-3 will also work for a^2-a=12

-3^2-(-3)=12
9+3=12
12=12


However, I still feel that B is correct.


Stmt 1:

a = 4 and a = -3

When a=4

sqrt(4^3-4^2-b)=7
sqrt(64-16-b)=7
sqrt(48-b)=7
b must be -1

when a=-3


sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be 67

so we don't know if a is -3 or 4. INSUFF



From stmt 2 we know that b=2 or b=-1

from the original statment we know that a and b are both integers.

so if b=2 there is no integer that solves the solution for a.

sqrt(a^3-a^2-2)=7
No integer solves this.

Now trying -1 as the solution.
sqrt(a^3-a^2-(-1))=7,
a=4 is the only integer that will make this correct.

So we know that a must be 4.


Maybe I am missing something on this.

dmateer25 wrote:

Maybe I am missing something on this.

On the same token, statement two gives you two values for a and -3 can not be the answer because we can not have negative number inside the root. So IMO , D is the correct answer here.

logitech wrote:

dmateer25 wrote:

Maybe I am missing something on this.

On the same token, statement two gives you two values for a and -3 can not be the answer because we can not have negative number inside the root. So IMO , D is the correct answer here.

Logitech,

Could you explain where I am going wrong with statement 1:

Stmt 1:

a = 4 and a = -3

When a=4

sqrt(4^3-4^2-b)=7
sqrt(64-16-b)=7
sqrt(48-b)=7
b must be -1

when a=-3


sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be -67

so we don't know if a is -3 or 4. INSUFF

cramya wrote:
Actually my approach was

a^2-a = 12
a(a-1) = 12



only possible when a=4 a-1=3 since its given a and b are integers

-3 ( -3-1) = 12 Cramya both values for a works in this case.

dmateer25 wrote:
when a=-3


sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be 67

so we don't know if a is -3 or 4. INSUFF

My friend, there is nothing wrong with your logic. ( Oh by the way b=-67).

In fact Statement 1 is INSUF since both a values satisfy the first question.

So B 'should' be the OA.

logitech wrote:

dmateer25 wrote:
when a=-3


sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be 67

so we don't know if a is -3 or 4. INSUFF

My friend, there is nothing wrong with your logic. ( Oh by the way b=-67).

In fact Statement 1 is INSUF since both a values satisfy the first question.

So B 'should' be the OA.

oops. Thanks for correcting me on the -67! I corrected it in my post.


This was a very tricky question!

Logitech,
I agree both of us missed that!


Dmateer,
You are absolutely correct. I made a mistake in not considering -3 as a possibility in stmt I (should have set it up as a quadratic like u).

The answer should be B) like u originally said!


Hope I am not messing up again

Should b not be -85?

sqrt(a^3-a^2-b) = 7

a^3-a^2-b=49

If a=-3

-27-((-3)^2) - b = 49

-27-9-b=49

-36-b=49
-b=85
b=-85


Regards,
Cramya

cramya wrote:Logitech,
I agree both of us missed that!


Dmateer,
You are absolutely correct. I made a mistake in not considering -3 as a possibility in stmt I (should have set it up as a quadratic like u).

The answer should be B) like u originally said!


Hope I am not messing up again

Should b not be -85?

sqrt(a^3-a^2-b) = 7

a^3-a^2-b=49

If a=-3

-27-((-3)^2) - b = 49

-27-9-b=49

-36-b=49
-b=85
b=-85


Regards,
Cramya

Cramya,

Yes, you are correct. I distributed the negative sign before squaring -3.

Let's just walk away from this question for god's sake!

Choose B and RUN!!

cramya wrote:I am getting D).

Whats the official answer?

The answer is B. I think because the first one can come out with a^3-a^2=48. a= 4 or -3.
the second one is b=-1, or 2.
B=-1 a=4
B=2, no values for a.

thus answer is B.