Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3

Question

Is √(x + y) an integer?

(1) x^3 = 64 
(2) x^2 = y – 3

pushpitkc · Answer

1. x^3 = 64
We can get the value of x = 4, but without the value of y - we can't
say whether √(x + y) is an integer.  (Insufficient)

2. x^2 = y – 3
If x = 1 and y = 4, then √(x + y) is not an integer
If x = 2 and y = 7, then √(x + y) is an integer  (Insufficient)
 
On combining the information in both the statements, we can clearly
get a unique value of x and y and √(x + y) is not an integer  (Sufficient - Option C)

Btw, are you sure this is from the OG - 2017(because if it is - it must have definitely been discussed)

vipulkhandelwal999 · Answer

I think OA should be B

because sqrt(x^2 + x + 3) will never be a perfect square for any integer value of x

push12345 · Answer

Take x as -3
U will get your answer

errorlogger · Answer

Hi  Bunuel ,

If x^3 = 64, x should be +4, -4
On Combining (1) (2):
We get √(x + y) = √17 or √(x + y) = 3

It the OA correct?

IanStewart · Answer

If x^3 = 64, then x can only be equal to 4. It can't be that x = -4, because (-4)^3 = -64, not +64.

You're surely thinking about even exponents: if you see equations like these, with a nonzero even exponent on one side and a positive number on the other, there will always be two solutions (you only have one solution when the right side is zero) :

x^2 = 25 --> x = 5 or -5
x^4 = 81 --> x = 3 or -3
x^20 = 1 --> x = 1 or -1

but if you see a similar equation with an odd exponent on the left side, there is always just one solution:

x^3 = 27 --> x = 3
x^5 = -32 --> x = -2
x^101 = 1 --> x = 1

Sonia2023 · Answer

chetan2u

Statement 1: x^3 = 64 -> Insufficient as we don't have the value of y.
Statement 2: x^2 = y – 3
x^2 +3 = y

Substituting in the question stem:
√(x + x^2 +3 -> √( x^2 +x + 3)
z= √( x^2 +x + 3)

Here solving for x will not give us anything because the determinant is negative - hence imaginary roots.  Since we are looking for whether x+y is a perfect square, it is not necessary that we need to have equal roots i.e. (x+a)^2 to get a perfect square?

If we combine the two - we will get √23 - not a perfect square. Hence sufficient

Can you comment on whether my interpretation (highlighted) is correct and how to think about this question?

Thank you

chetan2u · Answer

Hi

Highlighted part:  What you are calling as quadratic equation is a quadratic polynomial. 
It will become a quadratic equation if there is an 🟰 sign ahead.

You are trying to equal x^2+x+3 as 0 and then finding roots. But you do not know the value of x^2+x+3.

Solution

(1)  x^3 = 64 
Nothing about y
Insufficient

(2)  x^2 = y – 3 
You cannot get definite values of x and y. 
Insufficient

Combined
You know x and can get y from statement 2. Thus you can say for sure that whether  \sqrt{x+y}  is an integer or not. 
Sufficient

By getting into other aspects and not sticking to what has been asked in question, we would waste a lot of time on things that don’t help us in getting to solution.

But say you wanted to move ahead in statement 2 by substituting  y=x^2-3 
So,  x^2+x+3  could easily be equal to 4 with x as some value in decimal. Also, x=2 will give you x^2+x+3 as 9.
But  x^2+x+3  could also easily be equal to 5 with x as some value in decimal. 
Hence insufficient.

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Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3

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Is (x + y)^(1/2) an integer? (1) x^3 = 64 (2) x^2 = y – 3

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