Pre-Algebra Workbook 10: Radicals & Roots

Think of $\sqrt{81}$ as “the number which, when squared, equals $81$” Look for a whole number $n$ such that $n^2 = 81$

Find a number whose square is $81$

$$9^2 = 81$$

So

$$\sqrt{81} = 9$$

The correct choice is B. $9$.

Use the fact that $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ for positive $a$ and $b$ Take the square root of the numerator and of the denominator separately.

Separate the square root of the fraction:

$$\sqrt{\dfrac{49}{64}} = \dfrac{\sqrt{49}}{\sqrt{64}}$$

Evaluate each square root:

$$\sqrt{49} = 7, \quad \sqrt{64} = 8$$

So

$$\sqrt{\dfrac{49}{64}} = \dfrac{7}{8}$$

The correct choice is C. $\dfrac{7}{8}$.

Look for a perfect square factor of $50$ Write $50$ as $25 \cdot 2$ and use $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

Factor $50$ to find a perfect square factor:

$$50 = 25 \cdot 2$$

Rewrite the square root:

$$\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2}$$

Since $\sqrt{25} = 5$

$$\sqrt{50} = 5\sqrt{2}$$

The correct choice is B. $5\sqrt{2}$.

Look for the largest perfect square factor of $72$ Try $36$ $16$ $9$ or $4$ Then use $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

Factor $72$ using a perfect square:

$$72 = 36 \cdot 2$$

Rewrite the square root:

$$\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\sqrt{2}$$

Since $\sqrt{36} = 6$

$$\sqrt{72} = 6\sqrt{2}$$

The correct choice is C. $6\sqrt{2}$.

Like radicals (with the same number under the square root) combine like like terms in algebra. Add the coefficients and keep $\sqrt{5}$ the same.

Both terms have $\sqrt{5}$ so they are like radicals:

$$3\sqrt{5} + 2\sqrt{5} = (3 + 2)\sqrt{5} = 5\sqrt{5}$$

The correct choice is A. $5\sqrt{5}$.

First simplify $\sqrt{12}$ by factoring out a perfect square. Then check if the radicals are like terms so you can combine them.

First simplify $\sqrt{12}$

$$12 = 4 \cdot 3, \quad \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4}\sqrt{3} = 2\sqrt{3}$$

Now substitute back into the expression:

$$4\sqrt{3} - \sqrt{12} = 4\sqrt{3} - 2\sqrt{3}$$

Combine like radicals:

$$4\sqrt{3} - 2\sqrt{3} = (4 - 2)\sqrt{3} = 2\sqrt{3}$$

The correct choice is D. $2\sqrt{3}$.

Start by simplifying $\sqrt{28}$ using a perfect square factor. Then see if the radicals become like terms and can be combined.

Simplify $\sqrt{28}$

$$28 = 4 \cdot 7, \quad \sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}$$

Substitute back:

$$2\sqrt{7} + 3\sqrt{28} = 2\sqrt{7} + 3(2\sqrt{7}) = 2\sqrt{7} + 6\sqrt{7}$$

Combine like radicals:

$$2\sqrt{7} + 6\sqrt{7} = 8\sqrt{7}$$

The correct choice is B. $8\sqrt{7}$.

For a square, area $= s^2$ where $s$ is the side length. Set $s^2 = 64$ and take the square root to find $s$

For a square with side length $s$

$$s^2 = 64$$

Take the square root of both sides:

$$s = \sqrt{64} = 8$$

The side length is $8$ meters.

The correct choice is C. $8$ m.

Estimate $\sqrt{5}$ by noting that $2^2 = 4$ and $3^2 = 9$ Decide whether $\sqrt{5}$ is closer to $2$ or $3$ then compare with $2.2$

We know:

$$2^2 = 4, \quad 3^2 = 9$$

So $\sqrt{5}$ is between $2$ and $3$

Because $5$ is just a little larger than $4$ $\sqrt{5}$ is a little larger than $2$

A common approximation is $\sqrt{5} \approx 2.24$

Compare $2.24$ and $2.2$

$$2.24 > 2.20$$

so $\sqrt{5} > 2.2$

The correct choice is A. $\sqrt{5}$ is greater.

First simplify each square root by factoring out perfect squares: $18 = 9 \cdot 2$ and $8 = 4 \cdot 2$ Then see if the resulting radicals are like terms.

Simplify each radical separately.

For $\sqrt{18}$

$$18 = 9 \cdot 2, \quad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$$

For $\sqrt{8}$

$$8 = 4 \cdot 2, \quad \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$$

Now add the simplified expressions:

$$\sqrt{18} + \sqrt{8} = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}$$

The correct choice is D. $5\sqrt{2}$.