Is (6a + x^2)^(1/2) = a + x ?

Statement 1: Nothing mentioned about \(a\), hence the value cannot be calculated. \(x\) is either \(0\) or a negative no. Insufficient

Statement 2: nothing mentioned about x, hence value cannot be calculated. Insufficient

Combining 1 & 2, we know that x=0 or x<0 and a<0 so a+x<0. So RHS is always negative.
But LHS of the question stem is a square root which is always positive. Hence LHS is not equal to RHS. Sufficient

Option C

Chetan2u,
when the statement says "not a positive integer", is it safe to assume that it is either zero or a negative integer. I mean why do we rule out the fact that x is either a positive fraction or a negative fraction?
Do u understand what my concern is?
thank u in advance.

Hi afa13

This is my take on your query, other experts can pitch in their views.

All Real numbers are either Rational or Irrational

Rational number can be further divided into - Natural Number, Whole Numbers, Integers

Now Integers can be fractions also, you simply need to add 1 to the denominator. For eg. 2=2/1 (fraction), or 4/2 or 6/3 and so on. this means that every rational number can be written as a fraction.

Rational numbers can fall in more than one category - a Whole number can be an integer, similarly integers can be fractions.

So in this question if x is not a positive integer then we can safely assume that it will not be a positive fraction.

Dear Bunuel

This question triggers general question when I see something like Statement 1.

(1) x is not a positive integer.

Does it mean that x could be positive fraction or decimal such as 1/2 or 2.35 or any like numbers? or only mean that x< 0 or x =0?

IMO: nothing the statement limits the the option of positive fractions or positive decimals??

x is not a positive integer means that x is a positive number bur not an integer (1/2, \(\sqrt{2}\), ...) or x is a negative number or x is 0. So basically x is any number but 1, 2, 3, 4, ...

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Condition 1) & 2):
Since \(\sqrt{6a+x^2} ≥ 0\) and \(a < 0\), \(x≤0\), the left side is greater than or equal to 0 and the right side is less than 0.
Thus both sides never equal each other.
Both conditions together are sufficient.

Since this question is not one of key questions, we choose C.
Therefore, the answer is C.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

IMO C solve the question first 6a+x^2=a^2+2ax+x^2
6a=a^2+2ax
now come to statement
statement 1 x is not positive then x can be 0 or negative insufficient
statement 2 a is negative no information about x insufficient.
together we can say that not equal so sufficient.