6.5 Equations Involving Radicals BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 In this lesson, we will be solving equations where the variable

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6.5 Equations Involving Radicals

BobsMathClass.Com Copyright © 2010 All Rights Reserved.

32x 5 7 , 3yExamples: 4 x 2 , x 3 2x 1 4

In this lesson, we will be solving equations where the variable is part of the radicand (under the radical).

To solve equations of this form, we will need to utilize this next property of exponents:

n nIf a = b, then a b .

Let a and b be real numbers and n be a positive number.

Property of Exponents.

We will be utilizing this property to “cancel out” the radical. By using this property to solve equations with radicals, it will be necessary to perform the check to verify the solution. This process will produce answers which do not satisfy the original equation.

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From these examples, observe the result of the square of a square root equals the radicand. It is not even necessary to perform the multiplication. (The square “cancels out” the square root.)

What occurs when a square root is squared?Evaluate the following.

25

211

2x

5 5 25 5

11 11 121 11

2x x x x

Evaluate a few more.

21,261

22x 7

22x 9

1,261

2x 7

2x 9

1,261

2x 7

2x 9

What occurs when a cube root is cubed?

Evaluate the following.

33 2

33 5

33 x

3 3 3 32 2 2 8 2

3 3 3 35 5 5 125 5

33 3 3 3x x x x x

331261

Evaluate a few more.

33 2x 7

33 2x 9

From these examples, observe the result of the cube of a cube root equals the radicand. It is not even necessary to perform the multiplication. (The cube “cancels out” the cube root.)This idea of raising an nth root to the nth power will “cancel out” the radical. So if a radical has an index # = 5 such as:5 x 3 , raising the radical to the 5th power will “cancel out” the

radical.

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1. Isolate a radical expression on one side of the equation.Suppose the equation is x 3 4 0. To isolate the radical of the LHS (left hand side), 4 would be added to both sides.2. Use the property, if a = b, then an=bn. If the radical is a square root, square

both sides. If the radical is a cube root, cube both sides, if the radical is a 4th root, raise each side to the 4th power, if the radical is a 5th…, well you get the idea. In general, if the radical is an nth root, raise each side to the nth power.3. Solve the resulting equation. This will give preliminary solutions.

4. Check the solution or solutions. We may get results which do not satisfy the original equation. Yes, this step is necessary. If the solution checks, then it is a solution to the given equation.

The procedure for this lesson is not too complicated. This is not to say solving these equations is easy.

Procedure for solving Equations Involving Radicals.

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1. Isolate the radical on the LHS.

Solve x 3 4 0 and check yourExample solut. 1 ion.

Solution: x 3 4 0

+4 +4

x 3 4 2. Square both sides to “cancel out” the square root.3. Solve the equation.

2 2

x 3 4

x 3 16 -3 -3

x 13

4. Check the solution in the original equation.

Answer: 13Check: 13 3 4 0

16 4 0

4 4 0

0 0

Answer: 27

Your Turn Problem #1

Solve 2x 5 7 0 and check your solution.

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1. Isolate the radical on the LHS.

Solve 5x 1 6 0 and check yEx ouample r sol 2 u on. . ti

Solution:5x 1 6 0

-6 -6

5x 1 6 2. Square both sides to “cancel out” the square root.

3. Solve the equation. 2 2

5x 1 6

5x 1 36 4. Check the solution in the original equation.-1 -1

5x 35x 7 Actually, once a square root equals a negative

number, we can stop there because this is not possible. This procedure of raising both sides to power may give solutions which do no satisfy the original equation. These solutions are called extraneous solutions.

Check: 5(7) 1 6 0 36 6 0 6 6 0 12 0

This is not true. Therefore, there is no solution.

Answer:

Answer:


Solve x 5 2 0 .

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Solve 2x 4 x 2 and check yourEx sample olution(s). 3.

Solution:

1. Square both sides to “cancel out” the square root.

2. Solve the equation. Remember, when we have an x2, we need to get zero on one side, factor the binomial or trinomial, then set each factor equal to zero and solve. Right?3. Check the solutions in the original equation.

2 2

2x 4 x 2

2x 4 x 2 x 2 22x 4 x 4x 4

-2x+ 4 -2x + 420 x 6x 8

0 x 2 x 4

x 2 0 x 4 0 x 2 x 4

Check :

2( ) 42 2 2 4 4 0

0 0

x 2;

0 0

2( ) 44 4 2 8 4 2

4 2

x 4;

2 2

Answer: 2,4

Since both of the answers give a true statement, both numbers will be in the solution set.


Solve 2x 1 x 2 and check your solution(s). Hint: one of the answers will not check.

Answer: 5

(x=1 is an extraneous solution.)

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Solve 2 x x 3 and check your E sxampl olutie on .4 ). (s

Solution:1. We could divide by 2 on both sides to isolate the radical. But then we will have a rational expression on the RHS. That doesn’t sound too good. Let’s go ahead and square both sides.2. Solve the equation.

3. Check the solutions in the original equation.

2 2

2 x x 3

4x x 3 x 3 24x x 6x 9

20 x 10x 9

0 x 9 x 1

x 9 0 x 1 0 x 9 x 1

2 222 x 2 x

4x

Check:

Answer: 9x 9; 2 9 39

2 3 6 6 6

x 1; 2 1 31 2 1 2

2 2

Your Turn Problem #4Solve 4 x 5 x and check your solution(s).

Hint: one of the answers will not check. Answer: 25

(x=1 is an extraneous solution.)

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3 2 Solve 2x 5x 4 2 and check yourExamp solule t5 ion(s )..

Solution:1. Cube both sides to “cancel out” the cube root.2. Solve the equation.

3. Check the solutions in the original equation.

3

33 22x 5x 4 2 22x 5x 4 8

2x 3 x 4 0

2x 3 0 x 4 0 3

x2

x 4

22x 5x 12 0

Check :

3Answer: ,4

2

23 3 32 5 4 2

2 2

3x ;

2

39 15

2 4 24 2

39 15 8

22 2 2

33 168 2

2

23 2 4 5 4 4 2 x 4;

3 2 16 20 4 2

3 8 2


3 2Solve x 1 2 and check your solution(s). Answer: 3,3

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Solve x 4 x 1 1


In some equations you will need to square both sides more than once to clear all radicals.

SolvExample 6 x 2= 5e x+3

x 2 25 5 x 3 5 x 3 x 3

2 28 10 x 3

30 10 x 3

3 x 3

Solution: x = 5

2 2

Square box 2 5 x 3

x 2

th

5 x 3

si s

5

d

x

e

3

223 x 3

9 x 3

x 6 Please check.

Square both sides again.

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Pythagorean Theorem

Observe the right triangle to the right.A right triangle is a triangle where one of its angles is 90°. a and b are called legs of the triangle and c is called the hypotenuse.

b

ac

The Pythagorean Theorem is simply:2 2 2a b c

Example:

5

1213

This theorem holds true for the given triangle.

2 2 2

2 2 2

Using a b c , 5 12 13 25 144 169 169 169

ü We will now use the Pythagorean Theorem and the Square Root Property to find an unknown side. Note that when we use the Square Root Property, we place a ± symbol on the RHS. Since the length of a side can not be negative, we will not use the ± symbol.

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We can now make use of the Pythagorean Theorem to solve for d.

2 2 2d 12 16 2d 144 256 2d 112

d 112 4 7 10.6

Example 7. A 16-foot ladder resting against a house reaches a window-sill 12 feet above the ground. How far is the foot of the ladder from the foundation of the house? Express your answer in simplest radical form and to the nearest tenth of a foot.

Solution:

d = ?

Ladder

16 ft Wall 12ft

It would be appropriate to sketch the triangle made with the given information. d is the distance from the from the foot of the ladder to the foundation of the house.

The distance from the foot of the ladder tothe foundation is 4 7 or approximately 10.6 ft.

5 ft

Given the following right triangle, find the length of the missing side. Approximate to the nearest tenth.


11 ft

c=?

Answer: c 12.1 ft.

The End.B.R.6-10-07

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6.5 Equations Involving Radicals BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 In this lesson, we will be solving equations where the variable