Ch 3 (Powers and Radicals) PDF

Example 3-2 = 1 = 1 32 9 6 -4 10 10 (-1) =1 ; (-1) =1 ,(-y) =y 9 -5 9...

Example 3-2 = 1 = 1 32 9 6 -4 10 10 (-1) =1 ; (-1) =1 ,(-y) =y 9 -5 9 9 (-1) = -1 ; (-1) = -1 ; (-y) = - (y)  n n+2 n n 2 n n 2 (3 ) Remarks: 1) 3 ± 3 = 3.3 ± 3.1 = 3 ( 3 ± 1) = n ((3 ) n+2 n n+2 +n 2n +2 2) 3 x 3 =3 = 3 1) 2+5 =n → n=7 2) 3n=6 → n=2 3) 5-n=3 → n=2 3 2n 4) 2 x (2 ) = 2 7 → 3+2n=7 → 2n=4 → n=2 5) 0.1115x 101 = 0.1115x 10 n → n=1 -2 n 6) 735x 10 = 735 x 10 → n =-2 Exercise 1 page 40 7 5 7+5 x 12 1° x x = =x x -1 = 1x 3 -4 3-4 2° x x = x = 2-4-5 = x17 2 -4 -5 =x -7 3° x x x =x (23 ) (-23) -4 =x =x 9 (-2 ) (-3 ) 6 3 =x 5 4 = x 20 =x 10 8-4 4 4 = (-x) = (-x) =x ( ) 7-5 2 2 = (-x) = (-x) =x (The exponent is even) 9-6 3 ( ) 3 = (-x) = (-x) =-x Exercise 4 page 40 -2 -3 27 x 45 1° -5 4 = 36 x 45 6 -6 -15 2 (3 x7 ) 5 -3 ( 7 x 3) (3 x7 ) 7 6x 3 6 2° 2 3 2 6 3 = 4 6 6 3 (7 x5 ) 3 x7 (7 x5 ) 3 x7 -9 1 7 = = 6 6 7 6 6 16 3 x5 7 3 x5 7 = = = 3° -4 10 -16 3 2 x 10 (2 x5 ) 4° = 3 9 -2 -2 4 (4 x5 ) 5 x2 5 -20 3 10 10 -10 2 x5 x 2 x 5 2 x 5 13 =2x5 29 6 9 -2 -2 4 = -12 -18 2 (2 x5 ) x 5 x 2 x 5 2 x 5 x2x5  0 0.4. 5° =2 4 7 5 6° = = 3 The solution of equations in the form x 2= a If a < 0 (negative), then x 2= a has no solution in. 2 2 Example x = -4 has no real solution. (x is never negative) If a > 0 (positive),then x 2= a has 2 opposite real roots. Example x 2= 2 has 2 distinct real solutions: x= If a = 0, then x 2= 0 has 1 solution x = 0 (double root) Applications Solve for x the equations: 2 2 2 2 1) (2x-1) = - 9 2) (2x-1) = 7 3) (2x-1) = 9 4) (2x-1) = 0  2 1) (2x-1) = - 9 2 Impossible equation since (2x-1) is never negative. S= 2 2) (2x-1) = 7 → (2x – 1) = → 2x = 1 → x= (2 distinct real roots) S= 2 3) (2x-1) = 9 → (2x – 1) = ± 3 → → 2x = -3+1 or 2x = 3+1 → 2x = -2 2x = 4 → x = -1 x =2 S={-1 , 2} 2 4) (2x-1) = 0 → (2x-1)(2x-1)=0 1 → 2x-1=0 → x = 2 (Double root) Simplify the following fractions: 2 5 3 3 3 3 3 ( 32 – 1) 1) 3 – 3 4 2 = 3.3 – 3.1 = 2 2 = 3 2 2 2 3 – 3 3 ( 3 – 1) 3.3 – 3. 1 2n+2 2n 2n 2 2) 3 – 3 3.3 – 32n. 1 32n ( 32 – 1) n 3 n+2 n = n n = n 2 = 3 – 3 3.3 2 – 3. 1 3 ( 3 – 1) n n n n n n n 7 ( 1+2 ) n 3) 7 + 14 1+2 n n = 7.1 + 7.2 = n n = 7 – 14 n 7.1 n n – 7.2 7 ( 1 – 2 ) n 1– 2 n+2 n 2n+4 2n 4) 9 –9 3 – 3 3.3 – 3.1 2n 4 2n 2n 4 2n 2n+1 = = 3 (3 – 1) = -40 3 –3 2n 2n+1 = 3 – 3 2n 2n 1 2n 3. 1 – 3. 3 3 (1 – 3 ) 5 3 5 3 5) ( 3 ). (– 3) ( 3 ). [– (3) ] – (3) 8 -3 2 4 2 = 4 2 = 6 = ( 3). (–3) ( 3). (3) ( 3) Definition of a square root of a real Number a. x is a square root of a real number a if x 2= a. If a is negative then a has no square root since in this case x 2= a is impossible. Examples do not exist(they are not real №s) If a is positive then a has 2 square roots since in this case x 2= a has 2 opposite solutions The positive square root of a is. Example 1: Since then the square roots of 5 are..2 2.236 is the +ve square root of 5 Example2: Since (±2 ) = 4 then the square roots of 4 are ±2 but radical 4 is is the +ve square root of 4. The square root of 0 is 0, then Find x in each of the following cases: 1) → x=3 2) → x = -3 -x is positive when x is -ve 3) If x < 0 and Then x = -3 4) If x > 0 and Then x = 3 if x < 0 Rule: for x = 0 if x > 0 Write in terms of x the expression: f(x) 1) When x is negative f(x)= 3(-x)+ 2 (-x)- (-x)= -3x-2x+x = -4x 2) When x is positive f(x)= 3(x)+ 2 (x) - (x)= 3x + 2x - x = 4x 3) Deduce f(-3) and f(3) f(-3)= -4(-3)=12 and f(3) = 4(3)=12 The nth root of a real number * * * Notice that  : 4 x = 16 has 2 real solutions: x = and we say -2 and +2 are two 4th roots of 16  th n Root of a real number (n ,n≥2) n{2,3,4,5,…} Denoted by : n is the index , A is the radicand , Is the radical sign Existence and sign of If n is odd, then exists for any real number A In this case and A have the same sign. Examples: If n is even, then exists only when A is positive In this case is positive. Examples: More properties of = = ; ; Remarks: Simplify where all the variables have positive answers: 1) 3 6 7 3 3 a.b.c = 2 6 6 a.a.b. c. c = a.b.c ac 12 a.b.c 9 -5 = 12 8 -6 a. b.b. c.c 6 = a.b.c 4 -3 b.c 2) 6 4 a.b b.c = c3 5 3) 12 9 3 5 5 4 2 5 a.b.c = 10 2 a. a. b. b. c 3 = a.b a2.b c4 3 4 4 4 4) 12 9 3 = 12 8 3 3 2 a.b.c a.b. b. c = a.b b c3 10 10 5) 13 2 -10 10 -1 a3.b 2 a.b.c = 10 3 2 -10 a. a.b. c = a.c 10 = a a3.b 2 c  Simplify: 1) 3 4 2 2.3.5 = 2 4 2.2. 3. 5 = 2x3 2x5 5 2) 7 14 2 5 2 5 3.5.7 = 5 2 10 4 2 3.3.5. 5. 7 = 3x5 2 4 2 3 x5 x7 3 4 3 3) 12 9 3 = 2 X 5 x 11 2.5.11 4 2 2 4) 10 6 -8 5 3 -4 2 x3 2.3.5 = 2x3 x5 = 4 1 21 2.2.3. 3. 5 -4 2x3 = 2 3 3 3 5 5) 1080 = 3 3 2.3.5 = 2x3 5 6 6 6) 666 = 2.3.37 2 Simplify where all the variables have positive answers: 1) -4 a.b.c 5 -3 = -4 4 a.b. b c. c -4 -2 = a.b.c 2 -2 b c = b2 b.c a2c 2 -2 7 a.b.c = -2 6 =bc -1 3 ac 3 a.c 2) a. b c.c = c 5 3 -2a 3.c 5 5 b 3) 15 -10 25 a.b.c = a.b.c = b2 4 4 4 a 4 a2c 3 4) 4 18 -12 3 = 16 2 -12 3 a4 b -3 2 ac 3 = a.b.c a a.b.c = b 3 10 10 2 5 2 2 2 5) 20 5 -30 2 -3 10 5 a b a b = a 1 b a.b.c = ac b = 3 c = 3 3 c c 6) 3 -3 2 -4 2.3.5 = 2.3.3. 5.5 = 32 2x3x5 3 5 3 3 2 7) 11 -8 3 = 3 9 2 -9 3 3 -3 3 2 2x11 2 x5 2.5.11 2.2.5. 5.11 = 2x5 x11 2 x5 = 3 4 4 4 3 4 3 5 8) 432 = 2 X3 = 2 3  Exercise 8 page 40 1° 2° 3° 4° 5° 6° 7° 8° 11° 12° 6-1 6-2 Exercise 5 page 40 5 3 = x 3 2 = 5 -3 5 = 3 2 1 4 2 = 5 = 5 7 −2 −4 = 9−2 = 97 = 37 3 1 5 6 = (x y) 1 = ( xy ) 12  Choose the correct answer: 1) 2) b-a Rule: a-b Exercise 9 page 41 = = = = = = Exercise 10 page 41 1° 3 3 = 5 2° = 9 3 3° 1 = 2 4° 1 = 2 = -13 =- 13 ( ) – 25 5° = = 6° = = Exercise 11 page 41 1° 2° 3° 4° 5° 6° Page 41 3 False ,since 4x4x4 = 4 0 True, If a≠0, then a = 1 True 3 4 7 False, since 10 x10 = 10 8 8 8 False, since 5 x3 =15 2 False since , False since is true only if x≥0 False since if x < if x≤0  False, is ddefined only if x=0 False since True False, 13° No x verifies False since any –ve value of x satisfies the given equation False, it is an irrational № True 13 Page 42 > 2 > 2 Page 42 1° 2° = Then 3° Hence – = 6 – 6 =0 Properties of the nth root (A>0 , B>0 , m and n ) 7 7 Page 43 3 3 3 5 1° ( ) 3n 3n 2.2 + 2.1 23n ( 2 + 1) 2° = n n = n 2.2 – 2.1 2 (2 –1) 3° = But C Then Then 2 = 2n → n = 1 Exercise 19 page 43 23 23 24 23 1 24 2 1 1° A= 2 × 0.5 = 2 × 2) ( = 24 = 2 2 12 -2 2 12 -2 B= 9 × 32 (3 ) ( 25) 24 -10 3 = = 3 2 3 3 3 6 8 3 8 3 2 3 3 24 -10 143 3 2 3 6 6 = 1 = = 3 138 3 5 2 × 3 6 11 2 11 2 23 6 3 35 = 11 2 C= = = 2° 3° Radicals and absolute value A. Case of an even index which is a divisor of the exponent 1) = |-2| = 2 2) 3) = 4) Given a < 0 and b < 0 , the simplify : B. Case of an odd index which is a divisor of the exponent 1) =-2 2) 3) 4) Given a < 0 and b < 0 , the simplify : Find the followings: Use absolute value symbols only with even indices 5 5 1) ( 1- 3 ) = 1- 3 4 2) ( 1- 3 ) 4 =| 1 - 3 | = 3 – 1 3) 3 3 ( 1- 3 ) = 1- 3 7 4) ( 3 – 1) 7 = 3 – 1 5) 10 ( 3 – 1 ) 10 = | 3 – 1| = 3 – 1 Do the followings exist? 4 a) (1 - 3 ) Doesn’t exist 3 b) (1- 3 ) Exists Solve for x 1) x 5 = -1 2) x 6 = 12 6 3) (2-4x ) = 320 6 4) (2+4x) = -64 6 5) (2+4x) =64 6 6) Knowing that 4-2x < 0 ,solve (4-2x) =1 4 7) Knowing that 1+5x > 0 ,solve (1+5x) = 81 5 8) x4 =2 5 9) x3 =–2 Solve for x 5 → 5 x5 = 5 –1 → x = – 5 1 1) x = -1 → x = –1 6 12 → x = ± 6 12 2) x6 = 12 → 6 x 6 = 6 12 →|x| = 6 3) (2-4x ) = 320 → 6 2–4 x = 6 320 → |2 – 4x|= 6 320 → 2–4 x = ± 6 26 ×5 → 2–4 x = ±2 6 5 6 6 6 → 4x=2 2 5 → x=2 2 5 1 5 4 → x= 6 2 4) (2+4x) = -64 Impossible (A power of even exponent is never negative) 6 5) (2+4x) =64 → 6 (2+4 x) 6= 6 64 → | 2+ 4x |= 6 64 → 2+4 x = ± 6 2 6 → 4 x+2 = ± 2 → 4x = 0 or 4x = -4 → x=0 or x=-1 6) Given 4-2x < 0 ,solve (4-2x )6 =1 → 6 (4+2 x)6= 6 1 6 →| 4– 2x| = 1 → 4– 2x = ± 1 But given 4-2x< 0 → 4 – 2x = –1 → 2x=5 →  4 7) Given 1+5x > 0 ,solve (1+5x) =81 4 (1+5 x)4 = 4 81 →|1+5x |= 4 3 4 →1+5x = ±3 But given 1+5x > 0 → 1+5x = 3 → 5x = 2 → 5 5 5 5 8) x4 =2 → ( x 4 ) 4 =2 → x =2 5 4 4 4 4 2 → x= 25 → |x|= 2 → x =±2 4 2 5 5 5 5 9) 3 x3 =–2 → ( x 3 ) 3 = (–2) → x = – 2 5 x 3 = 3 -2 5 → x = –2 3 2 2 → x = –2 3 4 Exercise 7 page 40 1° 2° 3° 4° 5° 6° Outside part: Page 43 3 3 –ve 3 Exercise 14 page 42 For x < 0 , simplify each of the followings -ve(since the value of x is given –ve) 1° -ve (by given) 2° +ve (since x is given –ve then –x is +ve) 3° +ve(for any value of x and in particular if given x0 → E(x) is + ve then : + ve |–2x +y-4 | = –2x+y-4 2) Given E(x) = -2x+y-4 0 , and c < 0. Remarks: Use absolute value symbol only with even indices Examples:even –ve +ve +ve 1) odd 2) 3) –ve +ve +ve –ve +ve +ve 3 = 3 Absolute value and equations Consider on an axis x’ox the point M of abscissa x. Graphical interpretation of | | | Rule: |-z|=|z| Rule : | |= OM Rule: | | ≥ 0 Rule: |a-b|=|b-a| Solve |x|= 2 |x|=2 → | |= 2→ OM=2 → M is on M₁ or on M₂ →x = ±2 Rule: Examples : Solve for x 1) |x|= –1 → no solution (Since |x|≥ 0) 2) |x| = 2.5 → x= ±2.5 3) | 2x-3| = 5 → 2x-3 = ±5 → 2x = 3±5 → 2x = 8 or 2x = -2 → x = 4 or x = -1