Rules of Roots Combined Practice Problems & Solutions

Let's begin by calculating the numerical value of each of the roots in the given options:

$\sqrt{25}=5\\ \sqrt{16}=4\\ \sqrt{9}=3\\$We can determine that:

5>4>3>1 Therefore, the correct answer is option A

Let's start by recalling how to define a root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

Next, we will remember that raising 1 to any power will always yield the result 1, even the half power of the square root.

In other words:

$$$\sqrt{16}\cdot\sqrt{1}= \\ \downarrow\\ \sqrt{16}\cdot\sqrt[2]{1}=\\ \sqrt{16}\cdot 1^{\frac{1}{2}}=\\ \sqrt{16} \cdot1=\\ \sqrt{16} =\\ \boxed{4}$Therefore, the correct answer is answer D.

To solve the expression $\sqrt{1} \cdot \sqrt{25}$, we will use the Product Property of Square Roots.

According to the property, we have:

$\sqrt{1} \cdot \sqrt{25} = \sqrt{1 \cdot 25}$

First, calculate the product inside the square root:

$1 \cdot 25 = 25$

Now the expression simplifies to:

$\sqrt{25}$

Finding the square root of 25 gives us:

$5$

Thus, the value of $\sqrt{1} \cdot \sqrt{25}$ is $\boxed{5}$.

After comparing this solution with the provided choices, we see that the correct answer is choice 3.

Let's start by recalling how to define a square root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

Next, we remember that raising 1 to any power always gives us 1, even the half power we got from converting the square root.

In other words:

$$$\sqrt{1} \cdot \sqrt{2}= \\ \downarrow\\ \sqrt[2]{1}\cdot \sqrt{2}=\\ 1^{\frac{1}{2}} \cdot\sqrt{2} =\\ 1\cdot\sqrt{2}=\\ \boxed{\sqrt{2}}$Therefore, the correct answer is answer a.

In order to simplify the given expression, apply the following three laws of exponents:

a. Definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. Law of exponents for an exponent applied to terms in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

c. Law of exponents for an exponent raised to an exponent:

$(a^m)^n=a^{m\cdot n}$

Begin by converting the fourth root to an exponent using the law of exponents mentioned in a.:

$\sqrt{25x^4}= \\ \downarrow\\ (25x^4)^{\frac{1}{2}}=$

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

$(25x^4)^{\frac{1}{2}}= \\ 25^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}}$

We'll continue, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

$25^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}} = \\ 25^{\frac{1}{2}}\cdot x^{4\cdot\frac{1}{2}}=\\ 25^{\frac{1}{2}}\cdot x^{2}=\\ \sqrt{25}\cdot x^2=\\ \boxed{5x^2}$

In the final steps, we first converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in the reverse direction) and then calculated the known fourth root of 25.

Therefore, the correct answer is answer a.