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Mathematics LibreTexts

8.2: Simplifying Radical Expressions

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Let’s begin by comparing two mathematical expressions.

916=34=12

and

916=144=12

Note that both 916 and 916 equal 12. Hence, 916=916. Let’s look at another example.

49=23=6

and

49=36=6

Note that both 49 and 49 equal 6. Hence, 49=49. It appears that a pattern is forming, namely:

ab=ab

Let’s try an example on our calculator. First enter 23, then enter 23 (see Figure 8.2.1). Note that they produce the same result. Therefore, 23=23

fig 8.2.1.png
Figure 8.2.1: Note that 23=23

The above discussion leads us to the following result.

Multiplication property of radicals

If a0 and b0, then:ab=ab

Example 8.2.1

Simplify each of the following expressions as much as possible:

  1. 311
  2. 123
  3. 213

Solution

In each case, use the property ab=ab. That is, multiply the two numbers under the square root sign, placing the product under a single square root.

  1. 311=311=33
  2. 123=123=36=6
  3. 213=213=26

Exercise 8.2.1

Simplify: 28

Answer

4

Simple Radical Form

We can also use the property ab=ab in reverse to factor out a perfect square. For example:

50=252 Factor out a perfect square. =52 Simplify: 25=5

The expression 52 is said to be in simple radical form. Like reducing a fraction to lowest terms, you should always look to factor out a perfect square when possible.

Simple Radical Form

If possible, always factor out a perfect square.

Example 8.2.2

Place 8 in simple radical form.

Solution

From 8, we can factor out a perfect square, in this case 4.

8=42 Factor out a perfect square. =22 Simplify: 4=2

Exercise 8.2.2

Place 12 in simple radical form.

Answer

23

Sometimes, after factoring out a perfect square, you can still factor out another perfect square.

Example 8.2.3

Place 72 in simple radical form.

Solution

From 72, we can factor out a perfect square, in this case 9.

72=98 Factor out a perfect square. =38 Simplify: 9=3

However, from 8 we can factor out another perfect square, in this case 4.

=342 Factor out another perfect square. =322 Simplify: 4=2=62 Multiply: 32=6

Alternate solution

We can simplify the process by noting that we can factor out 36 from 72 to start the process.

72=362 Factor out a perfect square. =62 Simplify: 36=6

Although the second solution is more efficient, the first solution is still mathematically correct. The point to make here is that we must continue to factor out a perfect square whenever possible. Our answer is not in simple radical form until we can no longer factor out a perfect square.

Exercise 8.2.3

Place 200 in simple radical form.

Answer

102

The Pythagorean Theorem

An angle that measures 90 degrees is called a right angle. If one of the angles of a triangle is a right angle, then the triangle is called a right triangle. It is traditional to mark the right angle with a little square (see Figure 8.2.2).

fig 8.2.2.png
Figure 8.2.2: Right triangle ABC has a right angle at vertex C.

Right Triangle Terminology

  • The longest side of the right triangle, the side directly opposite the right angle, is called the hypotenuse of the right triangle.
  • The remaining two sides of the right triangle are called the legs of the right triangle.

Proof of the Pythagorean Theorem

Each side of the square in Figure 8.2.3 has been divided into two segments, one of length a, the other of length b.

fig 8.2.3.png
Figure 8.2.3: Proving the Pythagorean Theorem.

We can find the total area of the square by squaring any one of the sides of the square.

A=(a+b)2 Square a side to find area. A=a2+2ab+b2 Squaring a binomial pattern. 

Thus, the total area of the square is A=a2+2ab+b2.

A second approach to finding the area of the square is to sum the areas of the geometric parts that make up the square. We have four congruent right triangles, shaded in light red, with base a and height b. The area of each of these triangles is found by taking one-half times the base times the height; i.e., the area of each triangles is (1/2)ab. In the interior, we have a smaller square with side c. Its area is found by squaring its side; i.e., the area of the smaller square is c2.

The total area of the square is the sum of its parts, one smaller square and four congruent triangles. That is:

A=c2+4(12ab) Adding the area of the interior square and the area of four right triangles. A=c2+2ab Simplify: 4((1/2)ab)=2ab

The two expressions, a2+2ab+b2 and c2+2ab, both represent the total area of the large square. Hence, they must be equal to one another.

a2+2ab+b2=c2+2ab Each side of this equation represents the area of the large square. a2+b2=c2 Subtract 2ab from both sides. 

The last equation, a2+b2=c2, is called the Pythagorean Theorem.

The Pythagorean Theorem

If a and b are the legs of a right triangle and c is its hypotenuse, then:

a2+b2=c2

We say “The sum of the squares of the legs of a right triangle equals the square of its hypotenuse.”

Good hint: Note that the hypotenuse sits by itself on one side of the equation a2+b2=c2. The legs of the hypotenuse are on the other side.

Let’s put the Pythagorean Theorem to work.

Example 8.2.4

Find the length of the missing side of the right triangle shown below.

fig 8.2.4.png

Solution

First, write out the Pythagorean Theorem, then substitute the given values in the appropriate places.

a2+b2=c2 Pythagorean Theorem. (4)2+(3)2=c2 Substitute: 4 for a,3 for b16+9=c2 Square: (4)2=16,(3)2=925=c2 Add: 16+9=25

The equation c2=25 has two real solutions, c=5 and c=5. However, in this situation, c represents the length of the hypotenuse and must be a positive number. Hence:

c=5 Nonnegative square root. 

Thus, the length of the hypotenuse is 5.

Exercise 8.2.4

Find the missing side of the right triangle shown below.

Ex 8.2.4.png

Answer

13

Example 8.2.5

An isosceles right triangle has a hypotenuse of length 8. Find the lengths of the legs.

Solution

In general, an isosceles triangle is a triangle with two equal sides. In this case, an isosceles right triangle has two equal legs. We’ll let x represent the length of each leg.

fig 8.2.5.png

Use the Pythagorean Theorem, substituting x for each leg and 8 for the hypotenuse.

a2+b2=c2 Pythagorean Theorem. x2+x2=82 Substitute: x for a,x for b,8 for c.2x2=64 Combine like terms: x+x=2xx2=32 Divide both sides by 2

The equation x2=32 has two real solutions, x=32 and x=32. However, in this situation, x represents the length of each leg and must be a positive number. Hence:

x=32 Nonnegative square root. 

Remember, your final answer must be in simple radical form. We must factor out a perfect square when possible.

x=162 Factor out a perfect square. x=42 Simplify: 16=4

Thus, the length of each leg is 42.

Exercise 8.2.5

An isosceles right triangle has a hypotenuse of length 10. Find the lengths of the legs.

Answer

Each leg has length 52

Applications

Let’s try a word problem.

Example 8.2.6

A ladder 20 feet long leans against the garage wall. If the base of the ladder is 8 feet from the garage wall, how high up the garage wall does the ladder reach? Find an exact answer, then use your calculator to round your answer to the nearest tenth of a foot.

Solution

As always, we obey the Requirements for Word Problem Solutions.

  1. Set up a variable dictionary. We’ll create a well-marked diagram for this purpose, letting h represent the distance between the base of the garage wall and the upper tip of the ladder.

fig 8.2.6.png

  1. Set up an equation. Using the Pythagorean Theorem, we can write:82+h2=202 Pythagorean Theorem. 64+h2=400 Square: 82=64 and 202=400
  2. Solve the equation.h2=336 Subtract 64 from both sides. h=336h will be the nonnegative square root. h=1621 Factor out a perfect square. h=421 Simplify: 16=4
  3. Answer the question. The ladder reaches 421 feet up the wall. Using a calculator, this is about 18.3 feet, rounded to the nearest tenth of a foot.
  4. Look back. Understand that when we use 18.3 ft, an approximation, our solution will only check approximately.

fig 8.2.7.png

Using the Pythagorean Theorem: 82+18.32?=20264+334.89?=400398.89?=400

The approximation is not perfect, but it seems close enough to accept this solution.

Exercise 8.2.6

A ladder 15 feet long leans against a wall. If the base of the ladder is 6 feet from the wall, how high up the wall does the ladder reach? Use your calculator to round your answer to the nearest tenth of a foot.

Answer

13.7 feet.


This page titled 8.2: Simplifying Radical Expressions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold.

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