Math 11-1-Radical and Rational Expressions

Transcription

1 Math 11-1-Radical and Rational Expressions Math Absolute Value How to determine the expressions A positive number=the distance between the number zeroon the real number line. 8 = 8 =8 8 units 8 units Real number line How to determine absolute values in expressions Step-1: Perform all operations inside absolute sign Step-: Then replace the absolute with their equivalent value Step-3: Perform all remaining operations

2 Exp What is the value of ( 3 15 ) ( 5 3 ) ( 1 6 )( 4 7 ) ( 3 15 ) ( 5 3 ) ( 1 6 )( 4 7 )? Step-1: Step-: Step-3: ( 1 ) ( 5 )( 3 ) ( ) (1) () (5)(3) 6 15= 9 Exp What are the values of? (a) 9 14 Expressions Step-1 Step- Step-3 The value is (b) (c) 9 44 (d) (e) (f) Expressions Step-1 Step- Step-3 The value is (a) (b) (c) (d) (e) (f)

3 Math Radicals Radical Expressions Operations Adding - Subtracting Radicals Step-1: Simplify Step-: Then use distributive property Multiplying Radicals a x b y=(a b)( x y)=ab xy Dividing Radicals a x b y =( a b )( x y ) Rationalizing the Denominator of Radical Expressions Using the difference of squares (a b)(a+b)=a b Square Root Equations Solving then Verifying Verifying the solution to avoid extraneous roots such as x+3=0 none of positive numbers is 0 but x+3=0 ( x) =( 3) x=9 --> Verify the solution --> x+3=0 (9)+3=0 3+3=6 0 --> NO SOLUTION WITH EMPTY SET {}=

4 Exp Simplify =3 (3 3 ) (3 3 5)+5 ( 5) 6 ( ) (3 3) ( 3) 5+(5 ) 5 (6 4 4) = Exp. 1.. Simplify =(3 ) 18 10= (6 3 ) 5=36 Exp Simplify =( 6 6 ) 44 3 Exp Simplify ( ) ( 3+3 ) ) ( )=(3 3+3 ( 3 3 )( 3+3 ) = (3 5 5 ) ( 3+3 ) = (1 18) ( ) 6 ( ) = ( 6) ( ) ( 6) Exp Simplify (7+ 60) (5+ 8)+(3 9) (7+ 60) (5+ 8)+(3 9)= = = + 15

5 Exp Simplify = = = Exp If 88= x y, what are the value of x and y? 88= 3 3=( 3) =1 =x y x=1, y= Exp Simplify 16 16= =( 3) 3= 6 6 Exp Simplify = (3 7 3)( 5) = ( 5)( 5) ( ) 10 Exp Simplify ( ) =( 5) (3 7+ 3) (3 7) ( 3) = ( 5) (3 7+ 3) = 63 1 Exp If 3 7 3=P, what is the value of P? 3 6 3=(3 ) 3 3=(3 3) =18 = P P=18 Exp If =x y, what are the value of x and y? =( 35 7 )( 4 8 )=5 3= x y x=5, y=3

6 Exp Rationalize the denominator of =1 147 = ? Exp Algebraically solve this equation x+5=x+3? x+5=x+3 ( x+5) =(x+3) x+5=x +6x+9 x +5x+4=0 ( x+4)( x+1)=0 x= 1, 4 Verify the solutions x x+5= x = 1+3 4== Left side = Right side = 4+3 1=1 1 Left side is not equal to Right side The solution is x= 1 Exp Algebraically solve this equation x+1=x 1? Find extraneous root? x+1=x 1 ( x+1) =(x 1) x+1=4x 4x+1 4x 5x=0 ( x)(4x 5)=0 x=0, 5 4 Verify the solutions x x+1=x =(0) 1 1=1 1 Left side is not equal to Right side =( 5 4 ) 1 (5+4) 4 side = = 5 3 = 3 Left side = Right The solution is x= 5 4, the extraneous root is x=0

7 Exp Algebraically solve this equation x+=x+1? Find extraneous root? x+=x+1 ( x+) =( x+1) x+=x +x+1 x 1=0 ( x 1)( x+1)=0 x=1, 1 Verify the solutions x x+= x+1-1 ( 1)+=( 1)+1 0=0 Left side = Right side 1 (1)+=(1)+1 4= = Left side = Right side The solutions are x=1, 1, no extraneous roots. Exp Solve this equation x 9+ x=9? x 9+ x=9 x 9=9 x ( x 9) =(9 x) x 9=81 18 x+x 18 x=90 x= =5 ( x) =(5) x=5 Verify the solutions x x 9+ x= = =9 4+5=9 Left side = Right side Exp What is the quadratic form of x+11+ x=0 when you want it algebraically solved? x+11+=x x+11=x ( x+11) =(x ) x+11=x 4x+4 x 5x 7=0 The quadratic form should be x 5x 7=0 Exp What is the quadratic form of 1 x x 5=0? Can you algebraically solve it? 1 x x 5=0 1 x=x+5 ( 1 x) =(x+5) 1 x=x +10x+5 x +11x+4=0=(x+8)(x+3) x= 3, 8 Verify the solutions

8 x 1 x x 5=0-3 1 ( 3) ( 3) =0 Left side = Right side -8 1 ( 8) ( 8) = =6 0 Left side is not equal to Right side The quadratic form is x +11x+4=0 The solution is x= 3 Exp For any real number x, the equation of (5 x) =5 x is??? (a) Always true (b) Only true if x 5 (c) Only true if x 5 (d) never true The equation is true only if 5 x must be a non-negative real number 5 x 0 5 x x 5 (c) is correct.

9 Math Rational Expressions Simplifying Rational Expressions Step-1: Write it in factored form Step-: Identify common factor in numerator and denominator Step-3: Determine all non-permissible values Step-4: Reduce the common terms How to determine non-permissible values It is the value that make rational expression undefined such as x 3 x 1 x must not equal to 1, otherwise x 1=1 1=0 then it makes x 3 x 1 undefined. Adding - Subtracting Rational Expressions Step-1: Factor the numerators and denominators Step-: Simplify Step-3: Determine a common denominator and write each rational expression with the common denominator Step-4: Add or subtract the terms in the numerator Step-5: Identify non-permissible values and simplify further as required Multiplying - Dividing Rational Expressions Step-1: Factor the numerators and denominators Step-: Multiply the numerators and denominators Step-3: Dividing use reciprocal for multiplying Step-4: Identify non-permissible values and simplify further as required Algebraically Solving Rational Expressions Step-1: Factor the numerators and denominators, identify non-permissible values Step-: Reduce each rational expression Step-3: Identify the lowest common multiple of all factors of the denominators and multiply both sides of equation, reduce Step-3: Solve the resulting polynomial equation Step-4: Verify solutions against non-permissible values

10 Exp If x 5, simplify x 8x+15 x 10 Step-1: Write the expression in factored form --> x 8x+15 (x 3)( x 5) = x 10 (x 5) Step-: Identify the great common factor --> Step-3: Identify non-permissible values --> ( x 5) x 5 Step-4: Reduce the common factors --> x 8x+15 x 10 x 5) =(x 3)( = ( x 5) x 3 where x 5 Exp Simplify x+16 (x 5)(x 6x+8), what is the largest non-permissible value in this expression? Step-1: Write the expression in factored form --> x 8 (x 5)(x 6x+8) = (x 4) (x 5)(x )( x 4) Step-: Identify the great common factor --> Step-3: Identify non-permissible values --> ( x 4) x, x 4, x 5 Step-4: Reduce the common factors --> (x 5)(x ) x 8 (x 5)(x 6x+8) = (x 4) (x 5)(x )( x 4) where x, x 4, x 5, the largest non-permissible value is x=5 Exp Simplify 3 3x 6+6x Step-1: Write the expression in factored form --> 3 3x x) =3(1 6+6x 6(1+ x) Step-: Identify the great common factor --> 3 Step-3: Identify non-permissible values --> x 1 Step-4: Reduce the common factors --> 3 3x 6+6x =3(1 x) 6(1+x) = 1 x (1+ x) where x 1

11 Exp Simplify 5x +18x 8 5x x+8 Step-1: Write the expression in factored form --> 5x +18x 8 x+4) =(5x )( 5x x+8 (5x )( x 4) Step-: Identify the great common factor --> 5x Step-3: Identify non-permissible values --> x 5, x 4 Step-4: Reduce the common factors --> x 5, x 4 5x +18x 8 x+4) =(5x )( 5x x+8 (5x )( x 4) x+4 x 4 where Exp Simplify 54a 3 b 18a b Step-1: Write the expression in factored form --> 54a 3 b 18a b = (9 3)(a b)a b (9 )(a b) Step-: Identify the great common factor --> (9 )(a b) Step-3: Identify non-permissible values --> Step-4: Reduce the common factors --> a 0, b 0 a 0,b 0 54a 3 b 18a b = (9 3)(a b)a b = 3ab where (9 )(a b) Exp Simplify 4a 3a 8a Step-1: Write the expression in factored form --> 4a 3a = (4a)(1 8a) 8a (4 )(a) Step-: Identify the great common factor --> Step-3: Identify non-permissible values --> (4a) a 0 Step-4: Reduce the common factors --> 4a 3a = (4a)(1 8a) 8a (4 )(a) = 1 8a where a 0

12 Exp Identify non-permissible values in the following expressions x 15x x 3 9x (a) (b) (c) (d) x +4 4 x x x (a) Expressions x x +4 Identify non-permissible values x +4>0, x R There are not non-permissible values. (b) 15x 4 x >0 x ±, Non-permissible values are x=± 4 x (c) x 3 x x>0 x 0, Non-permissible value is x=0 (d) 9x x >0 x 0, Non-permissible value is x=0 x Exp Simplify the expression x 3x 10 x 6x x x 15 x 9 Step-1: Factoring --> x 3x 10 x 6x x x 15 ( x+)( x 5) x+3)( x 5) ( x 9 (x)(x 3) ( x 3)( x+3) Step-: Multiplying --> (x+)( x 5) (x)( x 3) (x 3)(x+3) (x+3)(x 5) = x+ x Step-3: Identify non-permissible values --> x 3,0,3,5 So x x 4 x x 30 x 5x x 5 x+ x where x 3,0,3, 5

13 Exp Simplify the expression a a 0 a 3a+ 4a 4 a 7a+10 Step-1: Factoring --> a a 0 a 3a+ 4a 4 a 7a+10 (a 5)(a+4) 4(a 1) (a 1)(a ) (a )(a 5) = Step-: Multiplying --> (a 5)(a+4) (a 1)(a ) 4(a 1) (a )(a 5) = a+4 4 Step-3: Identify non-permissible values --> a 1,,5 So a a 0 a 3a+ 4a 4 a 7a+10 a+4 4 where a 1,,5 Exp Simplify the expression a 4 + a+3 Step-1: Factor --> Step-: Simplify --> a 4 + a+3 ( This example doesn't need factoring, simplifying) Step-3: Determining a common denominator --> 4 Step-4: Adding --> a 4 +(a+3) = a () 4 + a+6 = 4 3a+4 4 Step-5: Identify non-permissible values and simplify further as required --> None So a 4 + a+3 = 3a+4 4 Exp Simplify the expression 3x 1 x+1 x+3 x Step-1: Factor --> 3x 1 x+1 x+3 x

14 Step-: Simplify --> 3x 1 x+1 x+3 x Step-3: Determining a common denominator --> x( x+1) Step-4: Subtracting --> 3x 1 x+1 x+3 = x (x)(3x 1) ( x+3)( x+1) (x+1)(x) = (3x x) ( x +x+3) (x+1)(x) 3x x+x x 3 = 4x 3x 3 (x+1)(x) ( x+1)( x) = Step-5: Identify non-permissible values and simplify further as required --> x 1,0 4x 3x 3 (x+1)(x) =(4x+3)(x 1) ( x+1)( x) So 3x 1 x+1 x+3 x = (4x+3)( x 1) (x+1)(x) where x 1, 0 Exp Simplify 8x 5 75x 15x 4x 3 Step-1: Factoring --> 8x 5 75x 15x 4x 3 =( )( x x x ) (3 5 5) x (3 5) x ( )(x x ) = Step-: Multiplying --> ( )( x x x ) (3 5 5) x (3 5) x ( )(x x ) = x 5 Step-3: Identify non-permissible values --> x 0 So 8x 5 75x 15x = x 4x 3 5 where x 0 Exp Simplify 64a 9a 18 a3 a 8a 3 Step-1: Factoring --> 64a a 9a 18 a3 = ( )(a ) 8a 3 9(a ) (a )(a a) ( )(a a ) =

15 Step-: Multiplying --> ( )(a ) (a )(a a) 9(a ) ( )(a a ) = 8a 9 Step-3: Identify non-permissible values --> a 0, So 64a a 9a 18 a3 = 8a 8a 3 9 where a 0, Exp Simplify a 5a+6 a 6a+8 a 3a a +a 0 Step-1: Factoring --> a 5a+6 a 6a+8 a 3a a +a 0 =(a )(a 3) (a )(a 4) (a)(a 3) (a+5)(a 4) = (a )(a 3) Step-: Multiplying --> (a )(a 4) (a+5)(a 4) = (a)(a 3) Step-3: Identify non-permissible values --> a 0,,3, 4 a+5 a So a 5a+6 a 6a+8 a 3a a +a 0 = a+5 a where a 0,,3, 4 Exp Simplify the expression 4a+8 + 9a 7 8a a Step-1: Factor --> Step-: Simplify --> 15a 40 15a + 1a+8 = (5)(3a 8) 3a 3a 8 3a + 1a+8 = 3a (3 5)(a) +1a+8 = 3a Step-3: Determining a common denominator --> 3a Step-4: Adding --> 3a 8+1a+8 = 4a 3a 3a 8 Step-5: Identify non-permissible values and simplify further as required --> a 0

16 So 4a+8 + 9a 7 8a a = 8 where a 0 Exp The expression is given as, what is the sum of all nonpermissible values? (Round the answer to the nearest tenth.) 3x+ (x +3x ) + x 5 10x 19x+6 3x+ (x +3x ) + x 5 10x 19x+6 = x (x+)(x 1) + x+1 (x 3)(5x ) The non permissible values are x=, 5, 1, 3 The sum of all non-permissible values is ( )+( 5 )+( 1 )+( 3 10 )=( )( 10 )+( 5 )( )+( 1 )( 5 5 )+( 3 )( 5 5 )=( 0 10 ) = Exp Step-1: Factor --> Simplify and find out all permissible values of this expression x +3x (x x 1) + x x 8x = x( x+3) (x+3)(x 4) + ( x x) x (x 4) = x +3x (x x 1) + x x 8x Step-: Simplify --> x x 4 + x x 4 = Step-3: Determining a common denominator --> x 4 Step-4: Adding --> x+ x x 4 = x x 4 Step-5: Identify non-permissible values and simplify further as required --> x 3, 0, 4 So x +3x (x x 1) + x x 8x = x x 4 where x 3, 0, 4 All non-permissible values are x= 3, 0, 4

17 Exp Algebraically solve the following equation x 4x+8 =1 x+4 x+4 Step-1: Factor the numerators and denominators --> x 4(x+7) =1 x+4 x+4 identify non-permissible values --> x 4 Step-: Reduce each rational expression Step-3: Identify the lowest common multiple of all factors of the denominators --> x+4 multiply both sides of equation, reduce --> x =( x+4) 4( x+7)= x+4 4x 3 x 3x 8=0 x ( x+4) = (x+4) ( 1 4(x+7) x+4 ) (x+4)= Step-3: Solve the resulting polynomial equation--> x 3x 8=0=( x+4)( x 7) x= 4,7 Step-4: Verify solutions against non-permissible values --> non-permissible values x 4 so x=7 Exp Find the value of a in 5 3a = 5 4a 6 8a Step-1: Factor the numerators and denominators --> 5 3a = 5 4a 6 8a identify non-permissible values --> a 0 Step-: Reduce each rational expression Step-3: Identify the lowest common multiple of all factors of the denominators --> 4a multiply both sides of equation, reduce --> ( 5 3a 6 ) (4a)= ( 5 4a 8a (5 3a)(4a)=(5 4a )(3)=0a 1a =15 1a 0a=15 Step-3: Solve the resulting polynomial equation--> 0a=15 a= 15 0 = 3 4 ) (4a) Step-4: Verify solutions against non-permissible values --> non-permissible values a 0 so a= 3 4

18 Exp Jack travels by plane,340 km and then by car for another 350 km to reach the local town in 1 Manitoba. The speed of the car is of the plane. Travelling by plane is 1 hour shorter than by car. How 9 long does it take Jack travelling by car? Let x = the speed the car in km/hr, y = the speed the plane in km/hr The speed of the car is 1 9 of the plane --> x= 1 9 y -----(1) The time of travelling by plane = 340 y, the time of travelling by car = 350 x Travelling by plane is 1 hour shorter than by car --> 350 x 340 = () y From (1) --> x= 1 8 y y=8x -----(3) Replace y in (3) at () --> 350 x =1 y x 340 9x =1 Step-1: Factor the numerators and denominators --> 350 x 340 9x =1 identify non-permissible values --> x 0 Step-: Reduce each rational expression Step-3: Identify the lowest common multiple of all factors of the denominators --> 9x multiply both sides of equation, reduce --> ( 350 x 340 9x ) (9x)=(1)(9x) (350)(9) 340=9x 9x=810 Step-3: Solve the resulting polynomial equation--> 9x=810 x= =90 Step-4: Verify solutions against non-permissible values --> non-permissible values x 0 so The speed of the car is x=90 km/hr Replace x in (1) --> x= 1 9 y 90=1 9 y y=90 9=810 The speed of the plane is y=810 km/hr

19 How long does it take Jack travelling by car? --> 350 x = =3 8 9 Jack travels by the car for hours. Exp True or False? Expressions True or False? (a) x x 8 ( x ) is simplified to x+4 where x (b) (x 9) (3 x x ) is simplified to x+3 where x 1,3 (a) (b) x x 8 = ( x+)(x 4) ( x ) (x ) (x 9) (3 x x ) x 3)(x+3) =( 1)( = (3 x) (3+x)(1 x) (1 x) Expressions True or False? (a) x x 8 ( x ) is simplified to x+4 where False --> Resulting is (x+)( x 4) ( x ) x where x (b) (x 9) (3 x x ) is simplified to x+3 False --> Resulting is (3 x) (1 x) where where x 1, 3 x 3, 1

20 Exp Norlan is in Physics lab today, he learns how to connect resistors in parallel. The physics teacher 1 give him a formula to calculate the resistance of resistors in parallel connection, = If R T R 1 R R T =5Ω, R 1 =0Ω 1 = = 1 R T R 1 R 5 = =( 1 R R 5 )( 3 3 ) 1 15 = = = 15 1 = R 15 ( 1 )( R 15 )=( R 15 )( R 15 ) R = 15 =7.5Ω Exp Given 4 m m = 3 m, what is the value of m? 4 m m = 3 m ( 4 m m ) ((m)(1+m))= ( 3 m ) ((m)(1+m)) ( 4 m ((m)(1+m))+ 1 1+m ((m)(1+m)) ) = ( 3 m ) ((m)(1+m)) 4(1+m)+m=3(1+m) 4+8m+m=3+6m 9m 6m=3 4 3m= 1 m= 1 3 Exp The fraction of is given, a number n is added to the numerator and subtracted from the denominator and give out the new fraction of 4 5 What is the value of n? n 90 n = 4 5 ( 36+n 90 n )(90 n)(5)=( 4 5 )(90 n)(5) (36+n)(5)=(4)(90 n) 180+5n=360 4n 5n+4n= n=180 n= n=0

21 Exp The distance from Vancouver to Prince George is 540 km, if I decrease the speed of driving by 15 km/hr, I will increase the time of driving by 7 min. What is the my current speed? Let x = the current speed of driving km/hr, t = the driving time (hr) By current speed, the driving time is t current = 540 x -----(1) if I decrease the speed of driving by 15 km/hr, the driving time is t new = 540 ( x 15) -----() I will increase the time of driving by 7 min --> t new t current = 540 x x = ( 1 x 15 1 x ) = ( 1 x 15 1 x ) = ( -----(3) 1 x 15 1 x ) ( x 15)( 5x 6 )=6 5 (x 15)( 5x 6 ) 90 ( ( x 15)(5x) ( x 15)(5x) x 15 x ) =( x 15)(x) 90 ((5x) (x 15)(5))=( x 15)(x) 90 (5x 5x+75 )=x 15x x 15x (90)(75)=0 x= b± b 4ac a x= 180, 150 x=90,75 x= ( 15)± ( 15) 4(1)( 90)(75) (1) Verify the correct speed --> x=90 --> replace x in (3) = = =36 5 6=36 30 = = 15± = 15±165 = x=75 --> replace x in (3) = = = = So my current speed is x=90 km per hr