What are those mysterious square roots that often confuse students and complicate their lives? The truth is that to understand them, we need to grasp the concept of the inverse operation.

What are those mysterious square roots that often confuse students and complicate their lives? The truth is that to understand them, we need to grasp the concept of the inverse operation.

When we solve an exercise like $5=25^2$, it's clear that $5$ times $5$ (that is, multiplying the number by itself) results in $25$. This is the concept of a power, or to be more precise, a square power, which to apply, we must multiply the figure or the number by itself.

**The concept of "square root" refers to the inverse operation of squaring numbers.**

That is, if we have $X^2=25$ and we want to find the value of $X$, what we need to do is perform an identical operation on both sides of the equation.

Choose the largest value

So, we have: $\sqrt{X^2} = \sqrt{25}$ and the result is $X=5$.

**Now let's explain in more detail the operation we have performed.**

- On the left side of the equation, the square root cancels out the squared power (the square root and the squared power are inverse operations, remember?).
- On the right side, we look for the figure or number that, when squared, gives us the result $25$.

There are two numbers that meet these requirements: $5$ and $-5$.

With that said, it's important to remember that **the square root of a number will always be** **positive**.

Therefore, to summarize, in the exercise: $X^2=25$ for example, we have two possible answers: $5$ and $-5$.

If we are given the mathematical expression $\sqrt{25}$ the only possible answer will be $5$.

- The only condition that must be met to find a square root is that
**the number under the square root must be****positive**.

**You cannot find the square root of a** **negative number**, that is, the expression $\sqrt{-25}$ is not correct and has no answer.

On the other hand,

**the result of a square root does not necessarily have to be an integer,**

that is, as long as the number under the square root is positive, we can find its square root.

**For example:** $\sqrt{89527}\approx 299.210$

Test your knowledge

Question 1

\( \sqrt{64}= \)

Question 2

\( \sqrt{36}= \)

Question 3

\( \sqrt{49}= \)

**Simple square roots are based on multiplication tables. Let's see some examples:**

$\sqrt{64}$

**Solution**: The square root, as we have already seen, is the inverse operation of squaring a number. Therefore, we must ask ourselves which number squared, or which number multiplied by itself, will result in $64$.

**Given that:** $8^2=64$

**the answer will be:** $\sqrt{64} = 8$

Do you know what the answer is?

Question 1

Solve the following exercise:

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

Question 2

\( 5+\sqrt{36}-1= \)

Question 3

\( \sqrt{441}= \)

$\sqrt{49}$

**Solution**: in this case, we also have to ask ourselves what number squared, or what number multiplied by itself, results in $49$.

Given that: $7^2=49$ the answer will be: $\sqrt{49} = 7$

$\sqrt{9}$

**Solution**: following the same logic, we see that $3^2=9$and, therefore, the answer will be: $\sqrt{9} = 3$.

Check your understanding

Question 1

\( \sqrt{225}= \)

Question 2

\( \sqrt{144}= \)

Question 3

\( \sqrt{121}= \)

In this section, we will apply what we have learned so far regarding square roots and see how we can use this information to solve algebraic exercises that include square roots.

**An important rule that we must remember when solving these types of exercises is that:**

Square roots should be solved first, that is, before any other mathematical operation that is outside of the square root itself.

$10 + \sqrt{81} =$

**Solution:**

First, we will solve the square root, as it comes before the addition that is outside of it.

**Thus, we obtain:**

$\sqrt{81} = 9$

**Then we continue with the rest of the exercise:**

$10+9=19$

Therefore, the result is $10 + \sqrt{81} = 19$

Do you think you will be able to solve it?

Question 1

\( \sqrt{100}= \)

Question 2

\( \sqrt{64}= \)

Question 3

\( \sqrt{36}= \)

$3 \times \sqrt{16} +8=$

**Solution:**

This exercise is a bit more complicated. Initially, we must solve the square root, as this operation precedes any other in the exercise.

Thus, we obtain: $\sqrt{16} = 4$

Afterwards, we should approach the exercise like any other math problem: $3 \times 4 +8$.

According to the order of operations in math, multiplication and division come before addition and subtraction. Therefore, the result is: $20$.

The answer to the exercise is: $3 \times \sqrt{16} +8=20$

$\sqrt{36} \div3+\sqrt{81}\times2=$

**Solution**:

Here we also need to solve the square roots first.

$\sqrt{36}=6$

$\sqrt{81}=9$

**We insert the numbers and solve according to the order of operations:**

$6\div3+9\times2=2+18=20$

**Therefore, the answer is:**

$\sqrt{36} \div3+\sqrt{81}\times2=20$

Test your knowledge

Question 1

\( \sqrt{16}= \)

Question 2

\( \sqrt{9}= \)

Question 3

Choose the largest value

Related Subjects

- Powers
- Division of Exponents with the Same Base
- Power of a Quotient
- Power of a Power
- Multiplying Exponents with the Same Base
- Exponent of a Multiplication
- Exponents for Seventh Graders
- The exponent of a power
- Order of Operations: (Exponents)
- Exponential Equations
- Multiplicative Inverse
- Order or Hierarchy of Operations with Fractions