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Lesson 1 Quadratic Equations and Functions PURPOSE This lesson gives you new techniques for solving quadratic equations, completing the square, using the quadratic formula, and rewriting polynomials as quadratics. OBJECTIVES After completing this lesson, you should be able to meet the following objectives: Solve quadratic equations by completing the square. Solve quadratic equations by using the quadratic formula. Determine the nature of the roots of a quadratic equation by using its discriminant. Solve equations in quadratic form. READING ASSIGNMENT Textbook, Chapter 7, Sections 7-1 through 7-4 COMMENTARY Section 7-1: Completing the Square (pages 307–309) Up to this point, you have only been able to solve a quadratic equation if it is factorable. For example: x 2 − x = 30 x 2 − x − 30 = 0 (x − 6)(x + 5) = 0 x−6=0 x = 6; x+5=0 x = − 5. Lesson 1 However, the main goal of algebra is to solve any equation given, so there must be a way to solve quadratic equations that are not factorable. One method is to complete the square. This is the process of working with an equation until it becomes an equation of the form (x + q) 2 = r. From there, the equation can be solved by taking the square root of each side: (2 x − 6) 2 = 8 q √ (2 x − 6) 2 = ± 8 √ (2 x − 6) = ± 2 2 √ 2x = 6±2 2 √ 6±2 2 x= 2 √ x = 3 ± 2. √ √ There are two solutions: 3 + 2 and 3 − 2. The process of completing the square takes the following steps. Let's start with the equation y 2 − 6 y + 5 = 3. 1. Isolate the constant y 2 − 6 y = − 2. 2. Add the square of 1 / 2 of the coefficient of the y term to both sides: µ ¶2 µ ¶2 1 1 y −6y+ 2 (6) = −2+ (6) . 2 2 This value must be added to both sides of the equation so that the value of the equation remains unchanged. 3. Factor the resulting trinomial: y 2 − 6 y + (3) 2 = − 2 + (3) 2 y2 − 6 y + 9 = − 2 + 9 y2 − 6 y + 9 = 7 (y − 3) (y − 3) = 7 (y − 3) 2 = 7. See the quantity squared? Now the equation can easily be solved: q √ (y − 3) 2 = ± 7 √ y−3= ± 7 √ y = 3 ± 7. √ √ There are two solutions: 3 + 7 and 3 − 7. Example 1: Solve −3 y + 6 = y 2 by completing the square. Lesson 1 Try the problem on your own and then check your work against mine: − 3 y + 6 = y2 y2 + 3 y = 6 µ ¶2 µ ¶2 1 1 y +3y+ 2 (3) = 6 + (3) 2 2 µ ¶2 µ ¶2 3 3 y2 + 3 y + =6+ 2 2 9 9 y2 + 3 y + =6+ 4 4 9 24 9 y2 + 3 y + = + 4 4 4 9 33 y2 + 3 y + = 4 4 µ ¶µ ¶ 3 3 33 y+ y+ = 2 2 4 µ ¶ 3 2 33 y+ = 2 4 sµ ¶ r 3 2 33 y+ = ± 2 4 µ ¶ √ 3 ± 33 y+ = 2 2 √ −3 33 y= ± ; 2 2 √ −3 ± 33 y= . 2 √ √ There are two solutions: −3 + 33 / 2 and −3 − 33 / 2. Factoring after adding the square of 1 / 2 of the x term is easy to do. Just use half of the x term in the factorization. So x 2 − 5 x + (5 / 2) 2 = 2 + (5 / 2) 2 becomes (x − 5 / 2) 2 = 33 / 4. The sign in the factorization is the same as the sign of the x term. Next, try factoring and simplifying the following equations. ¡1 ¢2 ¡1 ¢2 1. x 2 + 8 x + 2 (8) =6+ 2 (8) ¡1 ¢2 ¡1 ¢2 2. x 2 − 17 x + 2 (17) = −1+ 2 (17) ¡ 1 ¡ 3 ¢¢ 2 ¡ 1 ¡ 3 ¢¢ 2 3. x 2 − 3 x + 4 2 4 = −8+ 2 4 Answers 1. (x + 4) 2 = 22 Lesson 1 ¡ ¢2 2. x − 17 = 2 285 4 ¡ ¢2 3. x − 3 = − 503 8 64 Example 2: Solve x 2 − 9 x + 13 = − 8. Now use the factoring trick and solve this yourself; then check your answer and the process against the steps listed below: x 2 − 9 x + 13 = − 8 x 2 − 9 x = − 21 µ ¶2 µ ¶2 1 1 x −9x+ 2 (9) = − 21 + (9) 2 2 µ ¶2 µ ¶2 9 9 x2 − 9 x + = − 21 + 2 2 µ ¶ 9 2 84 81 x− = − + 2 4 4 µ ¶ 9 2 3 x− = − 2 4 sµ ¶ r 9 2 3 x− = ±i 2 4 µ ¶ √ 9 3 x− = ±i 2 2 √ 9 3 x= ±i ; 2 2 √ 9±i 3 x= . 2 When the coefficient of the term x 2 is not 1, completing the square takes one extra step. The coefficient must be factored out as a greatest common factor (GCF): 3 x2 − 8 x + 5 = 0 3 x 2 − 8 x = − 5. Divide every term on the left side by 3: µ ¶ 8 2 3 x − x = − 5. 3 Lesson 1 Divide the GCF over to the other side: 8 5 x2 − x= − 3 3 µ µ ¶¶ 2 µ µ ¶¶ 2 8 1 8 5 1 8 x2 − x + = − + 3 2 3 3 2 3 µ µ ¶¶ 2 µ ¶2 8 1 8 5 4 x − x+ 2 = − + 3 2 3 3 3 µ ¶2 4 5 16 x− = − + 3 3 9 µ ¶ 4 2 15 16 x− = − + 3 9 9 µ ¶2 4 1 x− = 3 9 sµ ¶ r 4 2 1 x− = ± 3 9 4 1 x− = ± 3 3 4 1 x= + 3 3 5 x = ; 3 4 1 x= − 3 3 3 x= or1. 3 When the solutions are rational, the problem is factorable. The problem 3 x 2 − 8 x + 5 = 0 can factor as (3 x − 5) (x − 1) = 0. It has solutions of 5 / 3 and 1. Example 3: Solve 4 x 2 − 2 x + 7 = 0. Try this problem yourself and then check your answer and the process. Lesson 1 4 x2 − 2 x + 7 = 0 4 x2 − 2 x = − 7 µ ¶ 1 4 x − x = −7 2 2 1 7 x2 − x= − 2 4 µ µ ¶¶ 2 µ µ ¶¶ 2 1 1 1 7 1 1 x2 − x + = − + 2 2 2 4 2 2 µ ¶ 1 2 7 1 x− = − + 4 4 16 µ ¶ 1 2 28 1 x− = − + 4 16 16 µ ¶ 1 2 27 x− = − 4 16 sµ ¶ r 1 2 27 x− = ±i 4 16 µ ¶ √ 1 ±3 i 3 x− = 4 4 √ 1 3i 3 x= ± 4 4 or √ 1 ±3i 3 x= 4 If a problem contains fractional coefficients, remove the fractions by multiplying the entire equation by the least common denominator (LCD). If a problem contains decimals, remove the decimal by multiplying by a multiple of 10. Example 4: Solve 1 x + .3 x 2 = 1 . 4 8 Multiply by 8, the LCD: 1 1 x + .3 x 2 = 4 8 µ ¶ 1 1 8 x + .3x =2 4 8 2 x + 2.4 x 2 = 1. Rearrange the terms: 2.4 x 2 + 2 x = 1. Lesson 1 Multiply by 10: 10 I2.4 x 2 + 2 x = 1M 24 x 2 + 20 x = 10. Finish it: 24 x 2 + 20 x = 10 µ ¶ 2 5 24 x + x = 10 6 5 5 x2 + x= 6 12 µ µ ¶¶ 2 µ µ ¶¶ 2 5 1 5 5 1 5 x2 + x + = + 6 2 6 12 2 6 µ ¶ 5 2 5 25 x+ = + 12 12 144 µ ¶2 5 60 25 x+ = + 12 144 144 µ ¶ 5 2 85 x+ = 12 144 sµ ¶ r 5 2 85 x+ = ± 12 144 µ ¶ √ 5 ± 85 x+ = 12 12 √ 5 85 x= − ± ; 12 12 √ −5 ± 85 x= . 12 Study Exercises Complete the odd-numbered written exercises 1–39 on pages 309–310 of the textbook and the odd mixed-review problems on page 310. Check your answers in the back of the textbook. Section 7-2: The Quadratic Formula (pages 311–313) The process of completing the square will always solve any quadratic equation. However, the process can be tedious and time consuming. The quadratic formula was derived by completing the square of the general quadratic equation a x 2 + b x + c = 0 . Because the quadratic formula was derived by completing the square, it can be used in place of completing the square. The derivation of the formula is on page 311 of your textbook. Lesson 1 In any quadratic equation, the coefficient of the x 2 term is a, the coefficient of the x term is b, and the constant term is c. These values then replace the a, b, and c in the quadratic formula: q¡ ¢ −b ± b2 − 4 a c x= . 2a So in the equation 3 x 2 + 2 x − 9 = 0, a = 3, b = 2, andc = − 9: q¡ ¢ −(2) ± (2) 2 − 4 (3) (−9) x= . 2 (3) The expression is then simplified to find the solutions: √ −2 ± (4 + 108) x= ; 6 √ −2 ± 112 x= ; 6 √ −2 ± 4 7 x= . 6 Bring out a GCF of 2 in the numerator; then simplify the fraction: ¡ √ ¢ 2 −1 ± 2 7 x= 6 √ −1 ± 2 7 x= . 3 Never reduce the fraction unless you bring out a GCF in the numerator. Then only cancel the GCF with the denominator. In the previous section, Example 3 solved 4 x 2 − 2 x + 7 = 0 by completing the square. One of the √ solutions was x = 1 ± 3 i 3 / 4. Let's use the quadratic formula to solve the equation and compare solutions. For the equation 4 x 2 − 2 x + 7 = 0, place the values into the quadratic equation: q¡ ¢ −b ± b2 − 4 a c x= 2a q¡ ¢ −(−2) ± (−2) 2 − 4 (4) (7) x= . 2 (4) Lesson 1 Now simplify the expression. You should obtain the same solutions found in the previous section: √ 2 ± (4 − 112) x= 8 √ √ 2 ± −108 2 ± 6i 3 x= → 8 8 ¡ √ ¢ ¡ √ ¢ 2 1 ± 3i 3 1 ± 3i 3 → . 8 4 Example 1: Solve 3 x (x + 2) = − 3.5. The quadratic formula cannot be used unless the equation is set equal to zero in the form a x 2 + b x + c = 0: a x2 + b x + c = 0 3 x (x + 2) = − 3.5 3 x 2 + 6 x + 3.5 = 0. Now find values of a, b, and c. Place them in the quadratic formula. Then simplify the expression: q¡ ¢ −(6) ± (6) 2 − 4 (3) (3.5) x= 2 (3) √ −6 ± (36 − 42) x= 6 √ −6 ± i 6 x= . 6 Six is not a GCF of the numerator. The solution cannot be simplified further. ± Example 2: Solve x 2 3 − 1 = x / 4. Multiply by the least common denominator: µ ¶ x2 x 12 −1= 3 4 4 x 2 − 12 = 3 x. Subtract over the 3x to set the equation equal to zero: 4 x 2 − 3 x − 12 = 0. Lesson 1 Finish solving using the quadratic formula: q¡ ¢ −(−3) ± (−3) 2 − 4 (4) (−12) x= 2 (4) √ 3± (9 + 192) x= 8 √ 3± 201 x= . 8 √ √ Example 3: Solve 5 x2 − 3 x + 2 5 = 0. Find values for a, b, and c and place them in the quadratic formula: q¡ ¡√ ¢ ¡ √ ¢¢ −(−3) ± (−3) 2 − 4 5 2 5 x= ¡√ ¢ . 2 5 Then simplify the expression: √ 3± (9 − 40) x= √ 2 5 √ 3 ± i 31 x= √ . 2 5 Rationalize the denominator: √ √ 3 ± i 31 5 x= √ •√ 2 5 5 √ √ 3 5 ± i 155 x= . 10 Example 4: Solve i x 2 + (1 − 2 i) x + (i − 1) = 0. In this equation a = i, b = (1 − 2 i), andc = (i − 1): q¡ ¢ −(1 − 2 i) ± (1 − 2 i) 2 − 4 (i) (i − 1) x= 2i √ −1 + 2 i ± ((−3 − 4 i) + (4 + 4 i)) x= . 2i Lesson 1 Simplifications of squared equations are: (4 + 4 i) − 4 (i) (i − 1) = − 4 i (i − 1) − 4 i2 + 4 i = 4 + 4 i; (−3 − 4 i) (1 − 2 i) 2 = (1 − 2 i) (1 − 2 i) 1 − 4 i + 4 i2 1 − 4i − 4 = − 3 − 4 i. √ −1 + 2 i ± 1 x= 2i 2i x= 2i x = 1; −1 + 2 i ± 1 x= 2i −2 + 2 i x= 2i 2 (−1 + i) x= 2i (−1 + i) i x= • i i ¡ ¢ −i + i 2 x= i2 (−i − 1) x= −1 x = i + 1. The solutions are 1 and i + 1. Example 5: Two numbers have a sum of 9 and a product of 17. What are the numbers? You already know that the two numbers have a sum of 9, so if you know one number, you can subtract it from 9 to find the other number. Thus, first number: x. Second number: 9 − x. Lesson 1 Product: x(9 − x) x (9 − x) = 17 9 x − x 2 = 17 x 2 − 9 x + 17 = 0 q¡ ¢ −b ± b2 − 4 a c x= 2a q¡ ¢ −(−9) ± (−9) 2 − 4 (1) (17) x= 2 (1) √ 9 ± (81 − 68) x= 2 √ 9 ± 13 x= 2 9 + 3.606 x= ; and 2 9 − 3.606 x= . 2 First number: x = 6.303; and x = 2.697. Second number: 9 − x = 2.697; and 9 − x = 6.303. The numbers are 6.303 and 2.697. Example 6: A box with height of (x + 2) cm has a square base with side x cm. A second box has a height of (x + 1) cm and a square base of (x − 1) cm. Both boxes have equal volume. Find the value of x. Volume of first box: x 2 (x + 2). Volume of second box: (x − 1) 2 (x + 1). Lesson 1 Volume is found by multiplying length, width, and height: x 2 (x + 2) = (x − 1) 2 (x + 1) x 3 + 2 x 2 = Ix 2 − 2 x + 1M (x + 1) x3 + 2 x2 = x3 − x2 − x + 1 3 x2 + x − 1 = 0 q¡ ¢ −(1) ± (1) 2 − 4 (3) (−1) x= 2 (3) √ −1 ± (1 + 12) x= 6 √ −1 ± 13 x= 6 −1 + 3.606 x= 6 x = .434; and −1 − 3.606 x= 6 x = − .768. The dimensions of the boxes with −.768 as the value of x would be as follows: x + 2 = 1.232 x = − .768; x + 1 = .232; x − 1 = − 1.768. Because there are negative dimensions, −.768 cannot be a value of x. The dimensions of the boxes with .434 as the value of x would be as follows: x + 2 = 2.434 x = .434; x + 1 = 1.434; x − 1 = − .566. Because there are negative dimensions, .434 cannot be a value of x. This problem has no solution. Study Exercises Complete the odd-numbered written problems 1–39 on pages 313–314 of the textbook and the odd word problems 1–13 on pages 314–315. Check your answers in the back of the textbook. Section 7-3: The Discriminant (pages 317–319) It is possible to determine if the solutions (also known as roots) of a quadratic equation will be the √ rational, irrational, or imaginary without actually finding√ solutions. Think about what values are √ rational, irrational and imaginary. Rational numbers are 16, 400, 6.25, etc. (numbers that are Lesson 1 √ √ √ perfect squares). Irrational numbers are 17, 300, 62.5, etc. (numbers that are not perfect √ √ √ squares). Imaginary numbers are −15, −6.25, −36 , etc. (square roots of negative numbers). The determining factor in all three sets of numbers deals with the square root. Therefore the value under the radical sign of the quadratic formula is used to determine the nature of the solutions. The expression b 2 − 4 a c is within the radical of the quadratic formula. It is so important that it is given a name: discriminant. If b 2 − 4 a c has a negative value, then the quadratic formula would take the square root of a negative number. If the value of b 2 − 4 a c is greater than or equal to zero, then the solutions are real. Real solutions can be either rational or irrational. If the value of b 2 − 4 a c is a perfect square, then it is rational; otherwise, it is irrational. Imaginary roots: b 2 − 4 a c < 0. Real rational roots: b 2 − 4 a c > 0 and b 2 − 4 a c is a perfect square. Real irrational roots: b 2 − 4 a c > 0 and b 2 − 4 a c is not a perfect square. There are always two solutions, −b + Áb 2 − 4 a c˜ ë 2 a and −b − Áb 2 − 4 a c˜ ë 2 a. If the roots are real, they q q È ˘ È ˘ will be two completely different roots and unequal. If the roots are imaginary, they are conjugates (3 − 4 i, 3 + 4 i). There is one more possibility for the value of b 2 − 4 a c. What would happen if the discriminant was zero? In the quadratic formula, the value of the radical expression is added to −b and subtracted from −b. So if the discriminant is zero, zero is added to −b and zero is subtracted from −b. The same root is found twice. Therefore, when b 2 − 4 a c = 0, there are real double roots. These rules only apply when the coefficients are real. These rules are listed on page 318 of your text. Let's try a few. ± Example 1: Given x 2 3 + 2 = 3 x, determine the nature of the roots. Rewrite the equation in the form a x 2 + b x + c = 0. The work will be easier if you remove the fraction as well: µ ¶ x2 3 +2 = 3x 3 x 2 − 9 x + 6 = 0. Now use the discriminant and make your determination: b2 − 4 a c (−9) 2 − 4 (1) (6) = 81 − 24 = 57. 57 is not a perfect square. The answer is real irrational roots. Lesson 1 Example 2: Given −2 x 2 − x = 6, determine the nature of the roots. Compare your answer to mine: − 2 x2 − x − 6 = 0 b2 − 4 a c (−1) 2 − 4 (−2) (−6) = 1 − 48 = − 47. The discriminant is negative. The answer is imaginary conjugates. Example 3: Given 5 x 2 = − 3 x + 2, determine the nature of the roots. First work this problem on your own and then compare your answer to mine: 5 x2 + 3 x − 2 = 0 b2 − 4 a c (3) 2 − 4 (5) (−2) = 9 + 40 = 49. 49 is a perfect square, thus the answer is real rational roots. √ Example 4: Given 5 x 2 + 2 10 x + 2 = 0, determine the nature of the roots. Compare your answer to mine: b2 − 4 a c √ 5 x 2 + 2 x 10 + 2 = 0 I2 10M − 4 (5) (2) = 40 − 40 √ 2 = 0. The answer is double real roots. Example 5: Find a value for k for which the equation k 2 x 2 − 6 x + 9 = 0 would have all of the following. (a) a real double root (b) two different real roots (c) imaginary roots Start by finding the discriminant of the given equation: b2 − 4 a c (−6) 2 − 4 Ik 2M (9) 36 − 36 k 2. Lesson 1 (a) If an equation has real double roots, the discriminant is zero. So set the expression for the discriminant equal to zero. Then solve for k: 36 − 36 k 2 = 0 36 = 36 k 2 1 = k2 ± 1 = k. (b) If an equation has two different real roots, the discriminant must be greater than zero. So set up an inequality stating that the discriminant is greater than zero: 36 − 36 k 2 > 0. Divide both sides by −36. Remember that dividing by a negative number switches the inequality sign: −36 Ik 2 − 1M > 0 k2 − 1 < 0 (k − 1) (k + 1) < 0 k = 1; k = − 1. −1 < k < 1 (c) If an equation has imaginary roots, the discriminant must be less than zero. So set up an inequality stating that the discriminant is less than zero: 36 − 36 k 2 < 0. Divide both sides by −36: − 36 Ik 2 − 1M < 0 k2 − 1 > 0 (k − 1) (k + 1) > 0 k = 1; k = − 1. Lesson 1 k < − 1 and k > 1 Study Exercises Complete the odd-numbered problems 1–39 on page 320 of the textbook and the odd mixed-review problems on page 321. Check your answers in the back of the textbook. Section 7-4: Equations in Quadratic Form (pages 322–323) All of the equations in this section can be thought of as trinomials, with a little imagination. Then solve the equations as you would any other quadratic equation. ¡ ¢2 ¡ ¢ Example 1: Solve 3 y 2 − 6 − 7 y 2 − 6 − 6 = 0. ¡ ¢ Substitute the quantity y 2 − 6 with x, and the equation becomes: 3 x 2 − 7 x − 6 = 0. The new simpler equation can be factored: (x − 3) (3 x + 2) = 0. ¡ ¢ Now substitute back in y 2 − 6 for x: IIy 2 − 6M − 3M I3 Iy 2 − 6M + 2M = 0. Simplify the equation: Iy 2 − 9M I3 y 2 − 18 + 2M = 0 Iy 2 − 9M I3 y 2 − 16M = 0. Factor again: (y + 3) (y − 3) I3 y 2 − 16M = 0. Lesson 1 Solve: y+3=0 y = − 3; y−3=0 y = 3; 3 y 2 − 16 = 0 3 y 2 = 16 √ √ √ ± 16 3 ±4 3 y = √ •√ = . 3 3 3 √ The solutions are −3, 3, and ±4 3 / 3. √ Example 2: Solve 36 / z − 13 / z + 1 = 0. √ What might be replaced by x in this problem? Let's be bold and replace 1 / z with x. Would it be true √ that (1 / z) 2 would be 1 / z? Yes. So 1 / z is x 2. I will write the equation a little differently so that you can see the fractions: µ ¶ µ ¶ 1 1 36 − 13 √ + 1 = 0. z z √ Now replace 1 / z with x and 1 / z with x 2: 36 x 2 − 13 x + 1 = 0. Factor: (9 x − 1) (4 x − 1) = 0. Solve: 9x−1 = 0 1 x= ; 9 4x−1 = 0 1 x= 4. √ Replace x with 1 / z: 1 1 √ = ; z 9 1 1 √ = . z 4 Lesson 1 Cross-multiply: √ 9= z; √ 4 = z. Square both sides to remove the radical: 81 = z; 16 = z. Any time you square during the process of solving an equation, you must check the solutions in the original problem. One or more of the solutions can be extraneous: 36 13 −√ +1=0 81 81 36 13 − +1=0 81 9 36 117 − +1=0 81 81 81 − +1=0 81 0 = 0; 36 13 −√ +1=0 16 16 36 13 − +1=0 16 4 36 52 − +1=0 16 16 16 − +1=0 16 0 = 0. Both 81 and 16 are solutions. Example 3: Solve x −4 − 10 x −2 + 9 = 0. Try this one yourself. Check your work and your solutions. x −4 − 10 x −2 + 9 = 0 Replace x −2with x: x 2 − 10 x + 9 = 0 (x − 9) (x − 1) = 0 x−9=0 x = 9; x−1=0 x = 1. Lesson 1 Replace x with x −2: 1 x −2 = x2 x −2 = 9 1 =9 x2 1 = 9 x2 1 = x2 9 1 ± = x; 3 x −2 = 1 1 =1 x2 1 = x2 1 = x2 ± 1 = x. Study Exercises Complete the odd-numbered problems 1–25 on page 324 of the textbook. Check your answers in the back of the textbook. It is almost time to take your first progress evaluation. To review, I suggest you do the following exercises: 1. Take the self-tests on pages 316 and 325 in your textbook. 2. Work the odd problems 1–7 in the Chapter Review on page 346 of the textbook. 3. Take the following practice quiz. Practice Quiz: Quadratic Equations and Functions You can check your answers against those listed at the end of this lesson. 1. Solve 2 x 2 + 6 x + 8 = 3 by completing the square. 2. Solve 4 x 2 + 2 x = − 3 using the quadratic formula. 3. Two numbers differ by 4. Their reciprocals add to 6. Find the numbers. 4. Determine the nature of the roots for the following quadratic equations: a. 3 x 2 + x − 7 = 0 b. −2 x 2 + x = − 3 c. 8 x 2 + 1 = 4 x 2 5. For what values of k will 3 x 2 = − k x − 12 have a double root? Lesson 1 ¡ ¢2 ¡ ¢ 6. Solve x 2 − 6 + 3 x 2 − 6 − 10 = 0. Answers This section provides answers to the practice quiz in Lesson 1. 1. 2 x2 + 6 x + 8 = 3 2 x2 + 6 x = − 5 2 Ix 2 + 3 xM = − 5 5 x2 + 3 x = − 2 µ ¶2 µ ¶2 1 5 1 x +3x+ 2 (3) = − + (3) 2 2 2 µ ¶ 3 2 5 9 x+ = − + 2 2 4 µ ¶ 3 2 10 9 x+ = − + 2 4 4 µ ¶ 3 2 1 x+ = − 2 4 sµ ¶ r 3 2 1 x+ = ± − 2 4 3 1 x+ = ± i 2 2 3 1 x= − ± i 2 2 2. 4 x2 + 2 x + 3 = 0 a = 4, b = 2, c = 3 q¡ ¢ −b ± b2 − 4 a c x= 2a q¡ ¢ −(2) ± (2) 2 − 4 (4) (3) x= 2 (4) √ √ √ −2 ± (4 − 48) −2 ± −44 −2 ± 2 i 11 x= → → 2 (4) 8 8 ¡ √ ¢ ¡ √ ¢ 2 −1 ± i 11 −1 ± i 11 x= → 8 4 Lesson 1 3. First number: x Second number: x − 4. First number's reciprocal: 1 x Second number's reciprocal: 1 . (x − 4) The reciprocals add up to 6: 1 1 + = 6. x (x − 4) Multiply by the LCD x (x − 4) to get (x − 4) + x = 6 x (x − 4): 2 x − 4 = 6 x 2 − 24 x 6 x 2 − 26 x + 4 = 0 q¡ ¢ p −b ± b2 − 4 a c −(−26) ± (−26) 2 − 4 (6) (4) x= → 2a 2 (6) √ √ 26 ± (676 − 96) 26 ± 580 → 12 12 √ ¡ √ ¢ √ 26 ± 2 145 2 13 ± 145 13 ± 145 x= → → 12 12 6 x = 4.17 x − 4 = .17; x = .16 x − 4 = − 3.84. 4. a. 3 x2 + x − 7 = 0 b 2 − 4 a c = (1) − 4 (3) (−7) (1) − 4 (3) (−7) = 1 + 84 1 + 84 = 85 Because 85 is not a perfect square, the roots are real and irrational. Lesson 1 b. − 2 x2 + x = − 3 b 2 − 4 a c = (1) − 4 (−2) (3) (1) − 4 (−2) (3) = 1 + 24 1 + 24 = 25 Because 25 is a perfect square, the roots are real and rational. c. 1 8 x2 + = 4x 2 µ ¶ 1 b 2 − 4 a c = (−4) 2 − 4 (8) 2 µ ¶ 1 − 4 (8) = 16 − 16 2 16 − 16 = 0 The discriminant is equal to zero. The roots are double real roots. 5. If it has a double root, the discriminant would have a value of zero: 3 x 2 = − k x − 12 3 x 2 + k x + 12 = 0 b 2 − 4 a c = k 2 − 4 (3) (12) = k 2 − 144. Set the value of the discriminant equal to zero and solve for k: k 2 − 144 = 0 k 2 = 144 k = ± 12. ¡ ¢2 ¡ ¢ 6. x 2 − 6 + 3 x 2 − 6 − 10 = 0 ¡ ¢ Replace x 2 − 6 with y: y 2 + 3 y − 10 = 0 (y + 5) (y − 2) = 0. Lesson 1 ¡ ¢ Replace y with x 2 − 6 : IIx 2 − 6M + 5M IIx 2 − 6M − 2M = 0 Ix 2 − 1M Ix 2 − 8M = 0 (x − 1) (x + 1) Ix 2 − 8M = 0 x−1 = 0 x = 1; x+1=0 x = − 1; x2 − 8 = 0 √ x = ± 2 2.