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Algebra - Positive/Negative Solutions of Squares and Square Roots

In today’s session we will try to clear your doubts regarding positive/negative solutions in case of squares and square roots. We will explain the reason behind each case and that will help you recall the fundamentals when you need to use them.

While preparing for any standardised test, you would have come across a discussion which says that x^2 = 4 has two roots: 2 and -2

but Sqrt(4) has only one value: 2


Now let’s try to understand why this is so.  


1. x^2 = 4


Basic algebra tells us that quadratics have two roots. Here, x can be either 2 or -2. Each when squared will give 4. 

x^2 - 4 = 0

(x + 2)*(x - 2) = 0

x is -2 or 2.


2. Sqrt(x) is positive only


Now this is odd, right? Sqrt(4) is 2 only. Why is that? Shouldn’t it be 2 or -2. After all, when we square either one of them, we get 4 (as discussed above). So, sqrt(4) should be 2 or -2. 


Here is the concept: Sqrt(x) denotes only the principal square root. x has two square roots - the positive square root (or principal square root) written as sqrt(x) and the negative square root written as - sqrt(x). When you take square root of 4, you get two roots: sqrt(4) and - sqrt(4) which are 2 and -2 respectively.


In a question, when you see sqrt(x), this is specifically the positive square root of the number. So sqrt(4) is 2 only. 


3. [Sqrt(x)]^2 = x


This is fairly straight forward. Since x has a square root, it must be non negative. When you square it, just the square root sign vanishes and you are left with x. 


4. Sqrt(x^2) = |x|


Now this isn’t intuitive either. Sqrt(x^2) should be simply x. Why do we have absolute value of x? Again, this has to do with the principal square root concept. First you will square x and then when you write sqrt, it is by default just the principal square root. The negative square root will be written as - sqrt(x^2). So irrespective of whether x was positive or negative initially, sqrt(x^2) will definitely be positive x. So we need to take the absolute value of x.


Quick recap with some examples:


  1. sqrt (9) = 3
  2. x^2 = 16 means x is either 4 or -4
  3. sqrt(5^2) = 5
  4. sqrt[(-5)^2] = 5
  5. [sqrt(16)]^2 = 16
  6. sqrt(100) = 10


To see this concept in action, we will take a very simple official problem.


Question 1: If x is not 0, then sqrt(x^2)/x = 


(A) -1

(B) 0

(C) 1

(D) x

(E) |x|/x


Solution: We know that sqrt(x^2) is not simply x but actually |x|.


So sqrt(x^2)/x = |x|/x


Depending on whether x is positive or negative, |x|/x will be 1 or -1 - we can’t say which one. Hence, there is no further simplification that we can do.

Answer must be (E).


Now that you are all warmed up, let’s look at a higher level question.


Question 2: Is sqrt[(x - 3)^2] = (3 - x)?


Statement 1: x is not 3

Statement 2: -x * |x| > 0



We know that sqrt(x^2) = |x|

So sqrt [(x - 3)^2] = |x - 3|


Then basically our question is:

Is |x - 3| = 3 - x?


Note that 3 - x is nothing but - (x - 3).


Is |x - 3| = -(x - 3)?


Recall the definition of absolute values. |a| = a if a >= 0 and -a if a < 0.

So, “is |x - 3| = -(x - 3)?” depends on whether (x - 3) is positive or negative. If (x - 3) is negative (or 0), then |x - 3| is equal to -(x - 3).

So the question boils down to 


“Is (x - 3) negative (or 0)?”


Statement 1: x is not 3


This means (x - 3) is not 0. But we don’t know whether it is negative or positive. Not sufficient. 


Statement 2: -x * |x| > 0


|x| is always non negative. So for the product to be positive, -x must be positive too. So x must be negative. If x is negative, x - 3 must be negative too. 

If (x - 3) is negative, |x - 3| is equal to - (x - 3). Hence, this statement alone is sufficient.


Answer (B)


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