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Math 101 Study Session Spring 2016 Test 5 Chapter 13 and Chapter 14 Sections 1, 2, and 3 April 27, 2016 Chapter 13 Section 1: Introduction to Radical Expressions A square root of a positive number x is a number whose square is x. √ The symbol is called a radical sign. The number and/or variable inside the radical sign is called the radicand. Square Roots of Perfect Squares √ √ √ √ 1 = 1 16 = 4 49 = 7 √ √ √100 = 10 √ √4 = 2 √25 = 5 √64 = 8 √121 = 11 9=3 36 = 6 81 = 9 144 = 12 If an integer is not a perfect square, its square root can only be approximated. For example, 2 is not a perfect square so its square root can only be approximated and is called an irrational number. √ 2 ≈ 1.4142135 . . . A radical expression is in simplest form when the radicand contains no factor greater than 1 that is a perfect square. The Product Property of Square Roots √ √ √ If a and b are positive real numbers, then ab = a · b. The square root of a2√ √ For any real number a, a2 = |a|. If a ≥ 0, then a2 = a. A variable or a product of variables written in exponential form is a perfect square when each exponent is an even number. 1 Chapter 13 Section 2: Addition and Subtraction of Radical Expressions We will use the distibutive property to simplify the sum or difference of radical expressions with the same radicand. √ √ √ √ 5 2 + 3 2 = (5 + 3) 2 = 8 2 √ √ √ √ 6 2x − 4 2x = (6 − 4) 2x = 2 2x Chapter 13 Section 3: Multiplication and Division of Radical Expressions Product Property of Square Roots √ √ √ a· b= a·b √ The Square of √ 2 a For a > 0, ( a) = a. The Quotient Property of Square Roots r √ a a If a and b are positive real numbers, the = √ b b Radical Expressions in Simplest Form A radical expression is in simplest form if: 1. The radicand contains no factor greater than 1 that is a perfect square. 2. There is no fraction under the radical sign. 3. There is no radical in the denominator of the fraction. Chapter 13 Section 4: Solving Equations Containing Radical Expressions An equation that contains a variable expression in a radicand is a radical equation. Property of Squaring Both Sides of an Equation If a and b are real numbers and a = b, then a2 = b2 . 2 Chapter 13 Section 5: Rational Exponents and Radical Expressions 1 Definition of a n 1 √ If n is a positive integer, then a n = n a. m Definition of a n √ 1 m If m and n are positive integers, and a n is a real number, then a n = n am . Cube Roots Fourth Roots Fifth Roots √Square√Roots √ √ √ 3 4 5 1 = 1 36 = 6 1 = 1 1 = 1 1=1 √ √ √ √ √ 3 4 5 4 = 2 49 = 7 8 = 2 16 = 2 32 = 2 √ √ √ √ √ 3 4 5 27 = 3 81 = 3 243 = 3 √ 9 = 3 √64 = 8 √ √ 3 4 16 = 4 81 = 9 64 = 4 256 = 4 √ √ √ √ 3 4 25 = 5 100 = 10 125 = 5 625 = 5 Chapter 14 Section 1: Solving Quadratic Equations by Factoring of Taking Square Roots An equation that can be written in the form ax2 + bx + c = 0 where a 6= 0, is a quadratic equation. A quadratic equation is also called a second-degree equation. To solve quadratic equations we will use the Principle of Zero Products “If a · b = 0 then a = 0 or b = 0” Principle of Taking√the Square Root of Each Side of an Equation If x2 = a, then x = ± a. Chapter 14 Section 2: Solving Quadratic Equations by Completing the Square A quadratic equation of the form x2 + bc = 0, x 6= 0, that cannot be solved by factoring can be solved by completing the square. The procedure is: 1. Write the equation in the form x2 + bx + c = 0. 2 b 2. Add to both sides of the equation x2 + bx = −c to get the equation x2 + bx + 2 2 2 b b = −c + 2 2 3 2 b 3. Factor the left hand side of the equation, so that the equation x + bx + = 2 2 2 2 b b b −c + becomes x + = −c + 2 2 2 2 4. Take the square root of each side of the equation. 5. Solve for x. Chapter 14 Section 3: Solving Quadratic Equations by Using the Quadratic Formula If ax2 + bx + c = 0, a 6= 0, then √ b2 − 4ac 2a or √ −b − b2 − 4ac x= 2a The quadratic formula is frequently written in the form √ −b ± b2 − 4ac x= 2a x= −b + Below are some examples for us to try with solutions at the end. 1. Simplify. √ −4 12a4 b7 2. Simplify. √ √ 7a 28ab2 − 7b 7a3 3. Simplify. p p √ 7x · 42x3 y · 6y 2 4. Divide. √ 7− 6 √ 5−2 6 5. Simplify. √ √ 6a5 b4 294ab4 6. Solve the equation and check your solution. √ √ 2x + 12 = 6 − 2 2x 4 7. Simplify. p 5 243x10 y 40 8. Simplify. 5 5 14x 2 y 4 5 −35xy 4 9. Solve by factoring. 8y 2 + 10y = 12 10. Solve by factoring. t2 + 5t − 14 = 0 11. Solve the equation by taking the square root. 3 (x − 6)2 = 12 12. Solve the equation by completing the square. x2 + 6x + 6 = 0 13. Solve the equation by using the Quadratic Formula 3x2 − 6x − 1 = 0 Solutions Solution to 1: √ √ −4 12a4 b7 = −4 4 · 3 · a4 · b6 · b √ √ = −4 4a4 b6 3b √ = −4(2)(a2 )(b3 ) 3b √ = −8a2 b3 3b Solution to 2: √ √ √ √ 7a 28ab2 − 7b 7a3 = 7a 7 · 4 · a · b2 − 7b 7 · a2 · a √ √ √ √ = 7a 4b2 7a − 7b a2 7a √ √ = 7a(2b) 7a − 7b(a) 7a √ √ = 14ab 7a − 7ab 7a √ = (14ab − 7ab) 7a √ = 7ab 7a 5 Solution to 3: p p p p 7xy · 42x3 y · 6y 2 = (7xy)(42x3 y)(6y 2 ) p = 42 · 42 · x4 · y 4 √ p p = (42)2 x4 y 4 = 42x2 y 2 Solution to 4: √ √ √ 7− 6 7− 6 5+2 6 √ = √ · √ 5−2 6 5−2 6 5+2 6 √ √ (7 − 6)(5 + 2 6) √ √ = (5 − 2 6)(5 + 2 6) √ √ √ √ 7(5) + 7(2 6) + (− 6)(5) + (− 6)(2 6) √ √ √ √ = 5(5) + 5(2 6) + (−2 6)(5) + (−2 6)(2 6) √ √ √ 35 + 14 6 − 5 6 − 2( 6)2 √ √ √ = 25 + 10 6 − 10 6 − 4( 6)2 √ 35 + 9 6 − 2(6) = 25 − 4(6) √ 35 − 12 + 9 6 = 25 √ − 24 = 23 + 9 6 Solution to 5: √ √ 6a5 b4 294ab4 r = r 6a5 b4 294ab4 6a5−1 r 6 · 49 a4 = √ 49 a4 =√ 49 2 a = 7 = 6 Solution to 6: √ √ 2x + 12 = 6 − 2x √ 2 √ 2 2x + 12 = 6 − 2x √ √ 2x + 12 = 6 − 2x 6 − 2x √ √ √ √ 2x + 12 = 6(6) + 6(− 2x) + (− 2x)(6) + (− 2x)(− 2x) √ √ √ 2x + 12 = 36 − 6 2x − 6 2x + ( 2x)2 √ 2x + 12 = 36 − 12 2x + 2x √ 12 = 36 − 12 2x √ 12 − 36 = −12 2x √ −24 = −12 2x −24 √ = 2x −12 √ 2 = 2x √ 22 = ( 2x)2 4 = 2x 2x 4 = 2 2 2=x Solution to 7: Note: 243 = 3 · 81 = 3 · 9 · 9 = 3 · 3 · 3 · 3 · 3 = 35 p p 5 243x10 y 40 = 5 35 x10 y 40 √ 10 40 5 = 35 · x 5 y 5 = 3x2 y 8 Solution to 8: 5 5 5 7 · 2x 2 −1 = 5 −7 · 5 −35xy 4 5 2 2x 2 − 2 = −5 3 2x 2 =− 5 14x 2 y 4 Solution to 9: 8y 2 + 10y − 12 = 0 7 Note: a · c = 8 · (−12) = −96. We seek to answer the question, “What two numbers multiply to give -96 AND add to give 10?” Note that 16·(−6) = −96 and 16+(−6) = 10. We will split the middle term 10y into 16y+−6y and use the factor by grouping method or the AC method. 8y 2 + 10y − 12 = 0 8y 2 + 16y + −6y − 12 = 0 8y 2 + 16y + (−6y − 12) = 0 8y (y + 2) + −6 (y + 2) = 0 (y + 2) (8y − 6) = 0 Using the principle of zero products, we will solve the two equations y + 2 = 0 8y − 6 = 0 3 6 Hence y = −2 and y = = . 8 4 Solution to 10: To solve this equation, we answer the question “What two numbers multiply to give -14 and add to give 5?” Note that 7 · −2 = −14 and 7 + (−2) = 5. t2 + 5t − 14 = 0 (t + 7) (t − 2) = 0 Using the principle of zero products, we will solve the two equations t+7=0 t−2=0 Hence, t = −7 and t = 2. Solution to 11: 3 (x − 6)2 = 12 3 (x − 6)2 12 = 3 3 2 (x − 6) = 4 q √ (x − 6)2 = ± 4 x − 6 = ±2 To find the two solutions, we solve the two equations x − 6 = 2 x − 6 = −2 8 Hence the two solutions are x = 8, x = 4. Solution to 12: Please not that we cannot solve this equation by factoring. We cannot find two numbers that multiply to give 6 and add to give 6. To complete the square, we first move the constant term 6 to the right hand side of the equation. x2 + 6x = −6 6 b Next we identify b (the number in front of x), b = 6. Then we divide b in half = = 3, 2 2 2 b = 32 = 9. We will add 9 to both sides of the equation. and square b over 2 2 x2 + 6x + 9 = 9 − 6 x2 + 6x + 9 is a perfect square. Two numbers that multiply to give 9 and add to give 6 are both 3. Hence, x2 + 6x + 9 = (x + 3)(x + 3). x2 + 6x + 9 = 3 (x + 3) (x + 3) = 3 (x + 3)2 = 3 q √ (x + 3)2 = ± 3 √ x+3=± 3 x = −3 ± √ 3 √ √ Hence, the two solutions are x = −3 + 3 and x = −3 − 3. Solution to 13: To solve the equation 3x2 − 6x − 1 = 0 we will use the quadratic formula with a = 3, b = −6, and c = −1. 9 √ b2 − 4ac 2a p −(−6) ± (−6)2 − 4(3)(−1) = 2(3) √ 6 ± 36 + 12 = √6 6 ± 48 = 6√ 6 ± 16 · 3 = 6√ 6±4 3 = 6√ 6 4 3 = ± 6 √6 2 3 =1± 3 √ √ 2 3 2 3 and x = 1 − . Hence, the two solutions are x = 1 + 3 3 x= −b ± 10