lesson 9 – radical functions and equations - math blog · lesson 9 – radical functions and...

Lesson 9 – Radical Functions and Equations

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In this lesson, we will learn some new properties of exponents, including those dealing with fraction exponents and notation.

Our function type for this lesson is Radical Functions and we will learn their characteristics, their graphs and we will solve their equations both graphically and algebraically.

Lesson Topics

Section 9.1: Operations on Radical Expressions

§ Addition/Subtraction § Multiplication § Division/Rationalizing the Denominator

Section 9.2: Roots, Radicals, and Rational Exponents

§ Square Roots § nth Roots § Rational Exponents

Section 9.3: Square Root Functions

• Key Characteristics of Square Root Functions Section 9.4: Cube Root Functions – Key Characteristics

• Key Characteristics of Cube Root Functions Section 9.5: Solving Radical Equations by Graphing Section 9.6: Solving Radical Equations Algebraically Section 9.7: Applications of Radical Functions


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Lesson 9 – Radical Functions and Equations - MiniLesson

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Lesson 9 - MiniLesson Section 9.1 – Operations on Radical Expressions

The first operations we will learn with radical expressions are addition and subtraction. To add and subtract radicals, you must have what are called “like” radicals. The examples below will make this clearer. Problem 1 WORKED EXAMPLE – Adding/Subtracting Radical Expressions Add or subtract the radical expressions as indicated and simplify. a)

!

5 2−3 2 = (5−3) 2=2 2

Start by factoring ! 2 from each term and then simplify to obtain the final result.

b)

!

5 3+2 2−4 3 = (5−4) 3+2 2= 3+2 2

The terms with ! 3 can be combined because they are like terms. The !2 2 term must stand on its own.

c)

!

−2 8 +2 2 = −2 4 i2+2 2= −2 4 2+2 2= −2(2) 2+2 2= −4 2+2 2= (−4+2) 2= −2 2

Start by simplifying ! 8 in the first term. Then factor

! 2 from each term and simplify to obtain the final result.

Problem 2 MEDIA EXAMPLE – Adding/Subtracting Radical Expressions Add or subtract the radical expressions as indicated and simplify. a)

!4 5 + 5 =

b)

!4 7 −3 11 −4 7 +5

c)

!4 5 −7 25 =


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Problem 3 YOU TRY – Adding/Subtracting Radical Expressions Add or subtract the radical expressions as indicated and simplify. a)

!6 3+2 3−5 3 =

b)

!5 2+3 3−4 2

c)

!3 27 −4 2+5 3 =

Product Property for Square Roots

For !!a≠0,!and!b≠0, ! a ib = a b and! a b = a ib

The product property above was used in a few of the addition/subtraction examples. The property also helps us understand how to multiply expressions with radicals. The examples below illustrate this idea. Problem 4 WORKED EXAMPLE – Multiplying Radical Expressions (1) Multiply the radical expressions as indicated and simplify. a)

! 3 3 = 3i3 = 9 =3

Use the Product Property to combine the radicals and then multiply the 3’s to get ! 9 which simplifies to 3.

b)

!

5 10 = 5i10= 50= 25i2= 25 2=5 2

Use the Product Property to combine the radicals and then multiply to get ! 50 . Note that we also use the Product Property once we factor the biggest perfect square out of 50 that we can. ! 25 = 5 to give the final, simplified result.


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Problem 5 WORKED EXAMPLE – Multiplying Radical Expressions (2) Multiply the radical expressions as indicated and simplify. a)

!

3 2− 5( ) =2 3− 3 5

=2 3− 15

Multiply ! 3 times each term in ( ). Then use the Product Property to combine ! 3 5 = 15 and obtain the final result.

b)

!

1+ 3( )2 = 1+ 3( ) 1+ 3( )=1+1 3+1 3+ 3 3=1+2 3+ 9=1+2 3+3= 4+2 3

Use the FOIL method to approach this problem multiplying each term in the first ( ) by each term in the 2nd ( ). Then simplify by combining like radicals and obtain the final result.

Problem 6 MEDIA EXAMPLE – Multiplying Radical Expressions (2) Multiply the radical expressions as indicated and simplify. a)

!

5 7 5−2 3( ) =

b)

!

−3+2 5( )2 =

Problem 7 YOU TRY – Multiplying Radical Expressions (2) Multiply the radical expressions as indicated and simplify. a)

!

3 6−5 2( ) =

b)

!

2+ 7( )2 =


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The last operation we will work with is division. To get there, we need to learn about conjugates.

Radical Conjugates

Given the radical expression !a+ b , its conjugate is !a− b

Given the radical expression !a− b , its conjugate is !a+ b

Radical Expression Radical Conjugate

!4+3 7

!4−3 7

!1− 2

!1+ 2

! 5 +7

! 5 −7

We will use the idea of conjugates as we work with radical division. Notice in the examples below that when conjugates are multiplied, the result is always an integer value. Problem 8 WORKED EXAMPLE – Multiplying Conjugates Multiply the radical expressions as indicated and simplify. a)

!

1+ 3( ) 1− 3( ) =1+ 3− 3− 3 3

=1+0− 9=1−3= −2

FOIL as we did in previous examples. Notice that each set of ( ) has the same terms different only by sign. Notice that the result is a real number that no longer contains a radical. Notice that our result is different than Problem 5 b) because the binomials are slightly different.

b)

!

2− 3( ) 2+ 3( ) = 4+2 3−2 3− 3 3

= 4+0− 9= 4−3=1

FOIL as we did before. Notice that this example has the “-“ part first but this does not matter. The result is still a real number with no radicals.


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If a radical division expression has radicals in the denominator, we need to continue simplifying. The method we use to remove radicals from the denominator is called rationalizing the denominator. Problem 9 WORKED EXAMPLE – Rationalizing the Denominator Divide the radical expressions as indicated and simplify leaving the denominator free of radicals. a)

!

12= 1

2i22

= 22 2

= 24

= 22

Begin by recognizing there is a radical in the denominator. Then, multiply numerator and denominator by this radical. Notice that doing so will result in a whole number in the denominator and a final, simplified result.

b)

!

232

= 216i2

= 24 2

= 12 2

= 12 2

i22

= 22 4

= 22i2

= 24

Begin by simplifying the denominator to remove the ! 16 . Reduce the resulting fraction. Multiply numerator and denominator by ! 2 . Simplify to obtain the final result with no radical in the denominator.


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Problem 10 MEDIA EXAMPLE – Rationalizing the Denominator Divide the radical expressions as indicated and simplify leaving the denominator free of radicals.

a)

!15=

b)

!320

=

Problem 11 YOU TRY – Rationalizing the Denominator Divide the radical expressions as indicated and simplify leaving the denominator free of radicals. a)

!1300

=

b)

!218

=


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More complicated radical fractions contain denominators with multiple terms. In order to completely simplify, we still need to remove radicals from the denominator. To accomplish this, we now use the conjugates that were explored earlier in the lesson. Problem 12 WORKED EXAMPLE – Rationalizing the Denominator Divide the radical expression as indicated and simplify leaving the denominator free of radicals.

!

13+ 2

= 13+ 2

i3− 23− 2

= 3− 29−3 2+3 2− 2 2

= 3− 29− 4

= 3− 29−2

= 3− 27

Can!also!be!written!as:! 37 −27

Recognize that the radical in the denominator must be removed in order to completely simplify the expression. Multiply numerator and denominator by the conjugate of

!3+ 2which is !3− 2 . Perform the multiplication. Note that 0.227 is approximate form for this answer rounded to 3 decimal places.

Problem 13 MEDIA EXAMPLE – Rationalizing the Denominator Divide the radical expression as indicated and simplify leaving the denominator free of radicals.

!−2

5−2 3=


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Section 9.2 – Roots, Radicals, and Rational Exponents

Square Roots

The notation a is radical notation for the square root of a. The notation

12a is rational exponent notation for the square root of a.

On your TI 83/84 calculator, you can use the symbol to compute square roots.

Examples

a) 25 = 5 because 52 = 25

b) Note that the square and square root “undo” each other.

( )22144 144 and 144 144= =

c) 251 2 = 25 = 5

d) 1/264 ( 64)− = − is not a real number because there is no real number, squared, that will give !−64 .

Nth Roots, Radicals & Rational Exponents

an = a1 n , the nth root of a

! apq = a

pq

!aq( )p = a p

q

Examples

a) 4256256 4/14 == Calculator Entry: 256^(1/4)

b) ( ) 321872187 717 −=−=− Calculator Entry: (-2187)^(1/7)

c) 47.21515 313 −≈−=− Calculator Entry: -15^(1/3)

d) 6/16 )324(324 −=− is not a real number Calculator Entry: (-324)^(1/6)

e) ! 81( )34 = 81( )3 4 =27 Calculator Entry: (81)^(3/4)


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Problem 14 MEDIA EXAMPLE – Compute with Radicals and Rational Exponents Use your calculator to compute each of the following in the real number system. Round to two decimal places as needed. a)! 49 b) !81 3 c) ! −49 d) ! −83

e) !−253 2 f) ! −25( )3 2

g) ! 497 h) ! 1234 Problem 15 YOU TRY – Compute with Radicals and Rational Exponents Use your calculator to compute each of the following in the real number system. Round to two decimal places as needed. a) ! 36 b) ! 643

c) !163 2 d) ! −25( )1 2

e) !273( )4 f) ! 819


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Section 9.3 – Square Root Functions – Key Characteristics

Square Root Function – Key Characteristics

A basic square root function in the real number system has the form:

)()( xpxf =

where p(x) is a polynomial and 0)( ≥xp .

Domain: To determine the domain of f(x), you want to find the values of x such that 0)( ≥xp . Horizontal Intercept: To determine the horizontal intercept, solve the equation p(x) = 0. Write your answer as an ordered pair. Vertical Intercept: To determine the vertical intercept, evaluate f(0). Write your answer as an ordered pair.

Problem 16 WORKED EXAMPLE – Key Characteristics of Square Root Functions Graph 2)( −= xxf and determine the vertical intercept, horizontal intercept, and domain. To graph, input into Y1 the following: 2nd>X^2 then x-2) so that Y1= )2( −x . Graph on the standard window (Zoom 6) to get the graph below:

DOMAIN Solve 02 ≥−x to get 2≥x . Therefore the domain is 2≥x .

HORIZONTAL INTERCEPT Solve x – 2 = 0 to get x = 2. The horizontal intercept is (2,0).

VERTICAL INTERCEPT Determine f(0) = 220 −=− which is not a real number. So, there is no vertical intercept.

Switching the Inequality Sign when Solving Inequalities

Remember, when you are solving an inequality and you multiply or divide by a negative number, then you must change the direction of the inequality symbol.


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Problem 17 MEDIA EXAMPLE – Key Characteristics of Square Root Functions For each of the functions below, determine the domain, horizontal intercept, and the vertical intercept (if it exists). Then sketch an accurate graph of each. Use interval notation for the domain. Round to two decimal places as needed. a) xxf 412)( −= Domain of )(xf :

Horizontal Intercept:

Vertical Intercept:

b) !!g(x)= 2x −5 Domain of )(xg :

Horizontal Intercept: Vertical Intercept:

Problem 18 YOU TRY – Key Characteristics of Square Root Functions

Given the function !!f (x)= x −4 , determine the domain, the horizontal intercept and the vertical intercept (if it exists). Then sketch an accurate graph of each. Use interval notation for the domain. Round to two decimal places as needed.

Domain of )(xf :



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Section 9.4 – Cube Root Functions – Key Characteristics

Cube Root Function – Key Characteristics

A basic cube root function in the real number system has the form 3 )()( xpxf = ,

where p(x) is a polynomial.

Domain: The domain of f(x) is all real numbers. This can be written as !(−∞,∞) , ! , or stated as “all real numbers”. Horizontal Intercept: To determine the horizontal intercept, solve the equation p(x) = 0. Write your answer as an ordered pair. Vertical Intercept: To determine the vertical intercept, evaluate f(0). Write your answer as an ordered pair.

Problem 19 WORKED EXAMPLE – Key Characteristics of Cube Root Functions Graph 3 84)( −= xxf and determine the vertical intercept, horizontal intercept and domain. To enter f(x) into your calculator, first rewrite the radical as a rational exponent:

3/13 )84(84 −=− xx Graph on the standard window (Zoom 6) to get the graph below:

DOMAIN The domain is all real numbers.

HORIZONTAL INTERCEPT Solve 4x – 8 = 0 to get x = 2. The horizontal intercept is (2,0).

VERTICAL INTERCEPT

288)0(4)0( 33 −=−=−=f The vertical intercept is (0, –2).


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Problem 20 MEDIA EXAMPLE – Key Characteristics of Cube Root Functions For each of the functions below, determine the domain, horizontal intercept, and the vertical intercept (if it exists). Then sketch an accurate graph of each. Use interval notation for the domain. Round to two decimal places as needed. a) 3 1527)( xxf −= Domain of )(xf :

Horizontal Intercept:

Vertical Intercept:

b) 3 102)( −= xxg Domain of )(xg :


Problem 21 YOU TRY – Key Characteristics of Cube Root Functions Given the function 3 8)( += xxf , determine the domain, the horizontal intercept and the vertical intercept (if it exists). Then sketch an accurate graph of each. Use interval notation for the domain. Round to two decimal places as needed.

Domain of )(xf :



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Section 9.5 – Solve Radical Equations by Graphing

Solve Radical Equations by Graphing

• Let Y1 = one side of the equation, Y2 = other side of the equation. • Determine an appropriate window to see important parts of the graph. • Use the Graphing/Intersection Method to identify the intersection. • The x-value of the intersection is the solution to the original equation. • If the graphs do not intersect, then there is no solution to original the equation.

Problem 22 WORKED EXAMPLE – Solve Radical Equations by Graphing Solve the equation 4310 =− x graphically.

Let Y1 = x310 − Let Y2 = 4 Graph on the standard window (Zoom:6) then determine the intersection (seen below).

The solution is the x-value of the intersection which in this case is x = –2.


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Problem 23 MEDIA EXAMPLE – Solve Radical Equations by Graphing Solve the equations graphically. Include a rough but accurate sketch of the graph and intersection point. Mark and label the intersection. Round answers to two decimal places. a) 5123 =−x

Solution: ______________________

b) 1142541 =−+ x

Solution: ______________________ Problem 24 YOU TRY – Solve Radical Equations by Graphing

Solve the equation !! 8x +33 = 4 graphically. Include a rough but accurate sketch of the graphs and intersection point. Mark and label the intersection. Round answers to two decimal places.

Solution: ______________


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Section 9.6 – Solve Radical Equations Algebraically

Solving Radical Equations Algebraically

• Isolate the radical part of the equation on one side and anything else on the other side. • Sometimes you will have radicals on both sides. • Raise both sides of the equation to a power that will “undo” the radical (2nd power to

eliminate square root, 3rd power to eliminate cube root). • Solve for the variable. • Check your answer! Not all solutions obtained will check properly in your equation.

Problem 25 WORKED EXAMPLE – Solve Radical Equations Algebraically Solve the equation 4310 =− x algebraically. Because the radical is isolated, we can square both sides to remove the square root.

( )

16310 4310

431022

=−=−

=−

xx

x

Next, solve for x.

2 63 16310

−==−=−

xxx

VERY IMPORTANT! Check x = –2 in the original equation to be sure it works! Not all solutions obtained using the process above will check properly in your equation. If an x-value does not check, then it is not a solution to the original equation.

!

10−3(−2) = 410+6 = 416 = 44 = 4 ✔

x = –2 is the solution to this equation.


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Problem 26 MEDIA EXAMPLE – Solve Radical Equations Algebraically Solve the equations algebraically. Write your answers in exact form, then give the decimal approximation rounded to the nearest hundredth. Be sure to check your final result! a) 5123 =−x

b) 1142541 =−+ x

Problem 27 YOU TRY – Solve Radical Equations Algebraically Solve the equations algebraically. Write your answers in exact form, then give the decimal approximation rounded to the nearest hundredth. Be sure to check your final result! a) 20743 =−− x

b) !!5+ 3x −1 =7

c) 42452 =+x d) 64523 =−− x


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Problem 28 WORKED EXAMPLE – Solve Radical Equations – More Advanced Solve the equation algebraically and check graphically: 6x x+ = . Be sure to check your final result! Since the radical is isolated, square both sides to remove the square root. Then, isolate x.

€

( )

06 60

6 6

6

2

2

2

22

=−−

−−==+=+

=+

xxxx

xxxx

xx

What we now have is a quadratic equation. The easiest and fastest way to work with this problem is through factoring. You can also use the Quadratic Formula or graphing.

3 or 203or 02

0)3)(2(06 2

=−==−=+

=−+=−−

xxxx

xxxx

CHECK:

When x = –2

22 ?24

?262

−≠−=

−=+−

x = –2 does not check so is not a solution

When x = 3

33 ?39

?363

==

=+

x = 3 checks so is a solution.

Graphical Check: !!Y1= x +6, Y2= x Window: Standard (Zoom:6)

Using the Grahing/Intersection Method, we obtain a verified solution of x = 3.


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Problem 29 MEDIA EXAMPLE – Solve Radical Equations – More Advanced Solve the equation algebraically and graphically: xx =−+ 71 . Be sure to check your final result!

Problem 30 YOU TRY – Solve Radical Equations – More Advanced Solve the equation algebraically and graphically: 46 +=+ xx . Be sure to check your final result!


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Section 9.7 – Applications of Rational Functions

Problem 31 MEDIA EXAMPLE – Applications of Rational Functions

Distance to the horizon can be approximated by the function !!D h( ) = 3h

2 where h represents the

height of the observer above the earth in feet and D represents distance to the horizon in miles. Write your answers using complete sentences. Round answers to two decimal places as needed.

a) Determine the distance to the horizon from the top of your car (estimate 5 feet), the top of a 100-foot building and the top of an 8000-foot mountain.

b) If the horizon is 50 miles away, determine the height of the observer above the earth.


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Problem 32 YOU TRY – Applications of Rational Functions The speed S, in meters per second, of a tsunami depends on the depth d, in meters, of the ocean. The function model for !

S d( ) is:

!!S d( ) = 9.8d Write your answers using complete sentences. Round answers to two decimal places as needed.

a) Determine the speed of a tsunami when the ocean depth is 100 meters.

b) How deep is the ocean when a tsunami is moving at 100 meters per second?

lesson 9 – radical functions and equations - math blog · lesson 9 – radical functions and...

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