§7.6 2Rad Radical Eqns Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§7.6 2Rad

Radical Eqns

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot College Mathematics

1

Bruce Mayer, PE

BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt

Review § 7.6

MTH 55

 Any QUESTIONS About

• §7.6 → Radical Equations

 Any QUESTIONS About HomeWork

• §7.6 → HW-29


2

Bruce Mayer, PE


Radical Equations

 A Radical Equation is an equation in which at least one variable appears in a radicand.

 Some Examples:

4

5 x

  

1 and m

  m



9.

4

5 x

  

1 and m

  m



9.


3

Bruce Mayer, PE


Solve Eqns with 2+ Rad. Terms

1. Isolate one of the radical terms.

2. Use the Exponent Power Rule

3. If a radical remains, perform steps

(1) and (2) again.

4. Solve the resulting equation.

5. Check the possible solutions in the original equation.


4

Bruce Mayer, PE


Example



Solve

 SOLUTION x 3 x 2 1.

x 3 x 2 1 x 3 x 2 1

 x



3

  x 2 1



2 

 x



2

 x



3

2

  x



2



1

 x



2



1





2 

1

2 x 3

 x 2



2 x 2 1


5

4



2 x



2

Bruce Mayer, PE


Example



Solve x 3 x 2 1.

 SOLUTION

2

2

2

 x





 x



2

2



2

4

6

 x

2

 Check 6 by Inspection → 3−2=1 

 Thus The number 6 checks and it IS the solution


6

Bruce Mayer, PE


Example



Solve x

   

2 x



3.

 SOLN x

 x



2

 

2 2 x



3

  

4 2 x

 

2 x



3



2

 

2 x

  

4 2 x

 

2 x



3

One radical is isolated.

We square both sides.



4 2 x 3

4 2 x



3



2



( x



9

9)

2

Square both sides.

16(2 x



3)

 x

2 

18 x



81

32 x



48

 x

2 

18 x



81

0

 x

2 

14 x



33

0



( x



3)( x



11)

Factoring x



3 or x



11

Using the principle of zero products


7

Bruce Mayer, PE


Example



Solve x 2 2 2 x



3.

 Check : x = 3 x = 11 x

   

2 x



3

3 2 2

1 2 6 3

1 2 9

1 2 3



 x

   

2 x



3

1 1 2 2 2 ( 11



9 2 22 3

3 2 25

3 2 5

3



3



 The numbers 3 and 11 check and

8 are then confirmed as solutions.

Chabot College Mathematics Bruce Mayer, PE


Example



Solve x + 7 + – x + 6 = 5.

Start by isolating one radical on one side of the equation by subtracting

– x + 6 from each side. Then square both sides.

x + 7 + – x + 6 = 5 x + 7 = 5 – – x + 6 x + 7

2

= 5 – – x + 6

2 x + 7 = 25 – 10 – x + 6 + ( – x + 6)


9

Bruce Mayer, PE


Example



Solve x + 7 + – x + 6 = 5.

This equation still contains a radical, so square both sides again.

Before doing this, isolate the radical term on the right.

x + 7 = 25 – 10 – x + 6 + ( – x + 6) x + 7

2 x

– 24

= 31 – x – 10 – x + 6

= –10 – x + 6 x

– 12 = –5 – x + 6

( x

– 12) 2

= –5 – x + 6

2

Subtract 31 and add

Divide by 2.

x .

Square both sides again.


10

Bruce Mayer, PE


Example



Solve x + 7 + – x + 6 = 5.

This equation still contains a radical, so square both sides again. x

2 – 24

( x

– 12) 2

= –5 – x + 6

2 x + 144 = ( –5) 2 – x + 6

2

(

Square both sides again.

ab )

2

= a

2 b

2 x

2 – 24 x + 144 = 25 ( – x + 6 ) x

2 – 24 x + 144 = –25 x + 150 x

2

+ x

– 6 = 0

Distributive property

Standard form

( x + 3)( x

– 2) = 0


11

Factor.

Bruce Mayer, PE


Example



Solve

Now finish solving the equation.

( x + 3)( x – 2) = 0 x + 3 = 0 or x

– 2 = 0 x + 7 + – x + 6 = 5.

Zero-factor property x = –3 or x = 2

Finally CHECK for Extraneous Solutions


12

Bruce Mayer, PE


Example



Solve x + 7 + – x + 6 = 5.

Check each potential solution, –3 and 2, in the original equation.

If x = –3, then If x = 2, then x + 7 + – x + 6 = 5?

x + 7 + – x + 6 = 5?

–3 + 7 + –( –3 ) + 6 = 5?

2 + 7 + –( 2 ) + 6 = 5?

4 + 9 = 5?



The solution set is { −3, 2 }.

5 = 5

9 + 4 = 5?



5 = 5


13

Bruce Mayer, PE


The Principle of Square Roots

 Recall the definition of the

PRINCIPAL Square Root

 For any NONnegative real number n , If x 2 = n , then x

 n or x

  n .


14

Bruce Mayer, PE


Recall The Pythagorean Theorem

 In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then a 2 + b 2 = c 2

Hypotenuse c a

Leg


15 b

Leg

Bruce Mayer, PE


Example



Pythagorus

 How long is a guy wire if it reaches from the top of a 14 ft pole to a point on the ground 8 ft from the pole?

 SOLUTION d

8

Diagram

14


16 d

2 

14

2 

8

2 

196



64



260

• We now use the principle of square roots. Since d represents a length, it follows that d is the positive square root of 260: d



260 ft d



16.125 ft.

Bruce Mayer, PE


Isosceles Right Triangle

 When both legs of a right triangle are the same size, we call the triangle an isosceles right triangle.

If one leg of an isosceles right triangle has length a then c

2  a

2  a

2 c

2 

2 a

2 c

 a 2.

c a a

17 c

2




2 a

2

Bruce Mayer, PE


Lengths for Isosceles Rt Triangles

 The length of the hypotenuse in an isosceles right triangle is the length of a leg times 2.


18 a

45 o a 2 a

45 o

Bruce Mayer, PE


Example



Isosceles Rt. Tri.

 The hypotenuse of an isosceles right triangle is 8 ft long. Find the length of a leg. Give an exact answer and an approximation to three decimal places .

 SOLUTION

45 o

8 = a 2

8

2

 a a

8

4 2

 a .

( after Rationalizing Divisor).

45 o

19 a

2  a a

2 

8

2

2 a

2  2

8


Exact answer:

Approximation: a

 a



4 2 ft

5.657 ft.

Bruce Mayer, PE


30 °-60°-90° Triangle

 A second special triangle is known as a 30 °-60°-90° right triangle, so named because of the measures of its angles


20

Bruce Mayer, PE


Lengths for 30/60/90 Rt Triangles

 The length of the longer leg in a

30/60/90 right triangle is the length hypotenuse is twice as long as the shorter leg.

30 o

2 a a 3


21

60 o a

Bruce Mayer, PE


Example



30 °-60°-90° Triangle

 The shorter leg of a 30/60/90 right triangle measures 12 in. Find the lengths of the other sides. Give exact answers and, where appropriate, an approximation to three decimal places .

 SOLUTION

• The hypotenuse is twice as long as the shorter leg, so we have a 3

30 o

2 a c = 2


22 a = 2( 12 ) = 24 in .

60 o

12

Bruce Mayer, PE


Example



30 °-60°-90° Triangle

 SOLUTION

 The length of the longer leg is the length of the shorter a 3

30 o

2 a

60 o

12 b

 a 3 = 12 3 in.

Exact answer: c



24 in., b



12 3 in.

Approximation: b



20.785 in.


23

Bruce Mayer, PE


WhiteBoard Work

 Problems From §7.6 Exercise Set

• 24, 34, 38, 48, 60

 Astronomical Unit

= Sun↔Earth

Distance

= 149 598 000 km


24

Bruce Mayer, PE


All Done for Today

The Solar

Star System


25

Bruce Mayer, PE


Chabot Mathematics

Appendix

r

2  s

2 

 r

 s

 r

 s



Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

–


26

Bruce Mayer, PE


Graph y = |x|

6

 Make T-table

y = |x | x

5

6

3

4

-1

0

1

2

-6

-5

-4

-3

-2

5

6

3

4

1

2

1

0

4

3

6

5

2

5

4

3

2

1

-6 -5 -4 -3 -2 -1

0

-1

0

-2

-3

-4

-5


27 y

1 2 3

-6 file =XY_Plot_0211.xls

Bruce Mayer, PE


4 5 x

6

-3 -2

2

1

-1

0

0

-1

-2

M55_§JBerland_Graphs_0806.xls

-3

5

4

3

1

4

5 y

2

-10 -8

3

3

2

-6

4

-4

5

1

-2

0

0

-1

-2

M55_§JBerland_Graphs_0806.xls

-3

-4

-5

2 4 6 8 x

10


28

Bruce Mayer, PE


§7.6 2Rad Radical Eqns Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§7.6 2Rad

Radical Eqns

The Solar

Star System

Chabot Mathematics

Appendix

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§7.6 2Rad Radical Eqns Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§7.6 2Rad

Radical Eqns

The Solar

Star System

Chabot Mathematics

Appendix

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