Squares and Square Roots | Algebra

In today’s post, we will try to clear up your doubts regarding positive and negative solutions in the case of squares and square roots. We will explain the reasons behind each case, which will help you recall the fundamentals when you need to use them. While preparing for a standardised test, you have probably come across a discussion that says $x^2 = 4$ has two roots, 2 and -2, while $\sqrt{4}$ 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 you 4.
$x^2 - 4 = 0$ and $(x + 2)\times(x - 2) = 0$ when $x$ equals -2 or 2.

2) $\sqrt}{x}$ is positive, only

Now this is odd, right? $\sqrt{4}$ must be 2. Why is that? Shouldn’t it be 2 or -2. After all, when we square both 2 and -2, 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 – $\sqrt{x}$ . Therefore, when you take the square root of 4, you get two roots: $\sqrt{4}$ and – $\sqrt{4}$ , which is 2 and -2 respectively.

On a GMAT question, when you see $\sqrt{x}$ , this is specifically referring to the positive square root of the number. So $\sqrt{4}$ is 2, only.

3) $(\sqrt{x})^2 = x$

This is fairly straightforward – 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 simply be $x$ – why do we have absolute value of $x$ , then? 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 $x$ . Therefore, we will need to take the absolute value of $x$ .

Here’s a quick recap with some examples:

• $\sqrt{9} = 3$

• $x^2 = 16$ means $x$ is either 4 or -4

• $\sqrt{(5^2)} = 5$

• $\sqrt{(-5^2)} = 5$

• $(\sqrt{16})^2 = 16$

• $\sqrt{100} = 10$

To see this concept in action, let’s take a look at a very simple official problem:

Question 1: If ${x}$ is not 0, then ${\frac{\sqrt{(x^2)}}{x} =}$

(A) -1

(B) 0

(D) ${x}$

(E) ${\frac{|x|}{x}}$

We know that $\sqrt{(x^2)}$ is not simply $x$ , but rather | $x$ |. So, $\frac{\sqrt{(x^2)}}{x} = \frac{|x|}{x}$ .

Depending on whether $x$ is positive or negative, $\frac{|x|}{x}$ will be 1 or -1, we can’t say which one. Hence, there is no further simplification that we can do, and our answer must be (E).

Now that you are all warmed up, let’s examine a higher-level question:

Question 2: Is ${\sqrt{[(x - 3)^2]} = (3 - x)}$ ?

Statement 1: ${x}$ is not 3

Statement 2: – ${x \times |x| > 0}$

We know that $\sqrt{(x^2)} = |x|$ , so $\sqrt{[(x - 3)^2]} = |x - 3|$ .

This means that our question is basically:

Is $|x - 3| = 3 - x$ ?

Note that $3 - x$ can also be written as – $(x - 3)$ .

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

Recall the definition of absolute values: $|a| = a$ if $a$ is greater than or equal to 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 our question now boils down to:

Is $(x - 3)$ negative (or 0)?

Statement 1: $x$ is not 3

This means we know that ( $x - 3$ ) is not 0, but we still don’t know whether it is negative or positive. This statement is not sufficient.

Statement 2: $-x \times |x| > 0$

| $x$ | is always non-negative, so for the product to be positive, “- $x$ ” must also be positive. This means $x$ 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, and our answer is (B).

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