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Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Real part Imaginary part a _ Solving Quadratic Equations by Formula bi KEY Complex Numbers All the quadratic equations solved to this point have had two real solutions or roots. In some cases, solutions involved a “double root,” but there were always two solutions. Study the quadratic equation 0 = x2 + x + 2. • Can this equation be factored in order to find the solutions or roots? No, because the polynomial cannot be factored. • Can the zeros be found from the graph of the function y = x2 + x + 2? Explain. No, because the graph of the function does not cross the x-axis. There are still two solutions to this quadratic equation. However, they are not real numbers. Instead, the solutions involve numbers that are imaginary. The imaginary number unit is designated as “i.” There are two ways that we define i. Words Symbols • “i” is the number you can square to get an answer of -1. • Or, “i” is the square root of -1 i 2 = -1 i= −1 Square roots of negative numbers can be simplified using “i”: − x = (−1) x = −1 × x = i x • If x is a non-negative real number, then When you have a negative number under radical, you can take the negative out as an “i.” a Example Problems: Simplify the following complex numbers by writing them in terms of i. A) 49 ©2012, TESCCC = 7i Check: ( 7i )2 = 49i2 = 49(-1) = -49 09/06/12 page 1 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 B) 121 C) = 17 42 D) 3 E) 11i = = 12 + 36 2 ©2012, TESCCC Check: ( 11i)2 = 121i2 = 121(-1) = -121 5i 3 i 6+3i 09/06/12 page 2 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Formula KEY A complex number is the sum of a real number and an imaginary number, usually written in the form a + bi. Write the following numbers in a + bi form, and identify the real and imaginary parts. Simplify first if necessary. F) 25 25 + 0i Real part = 25 Imaginary part = 0i 9 + 144 6 I) 1.5 + 2i Real part = 1.5 Imaginary part = 2i G) 25i 0 + 25i Real part = 0 Imaginary part = 25i H) 3 – 5i 3 – 5i Real part = 3 Imaginary part = -5i J) 3 + 5i – 7 + 8i K) (6 + 2i)(6 – 2i) -4 + 13i Real part = -4 Imaginary part = 13i 40 Real part = 40 Imaginary part = 0i The conjugate of a complex number has the same real 2 + 3i part but an opposite imaginary part. (For example, 2 3i is the complex conjugate of .) Complex Conjugates a + bi and a bi * When complex conjugates are added—or multiplied—the result will have an imaginary part of 0 i. Tell the conjugate of each complex number. Number Conjugate Number Conjugate L) 25 25 N) -3 + 6i -3 – 6i M) 25i -25i O) -12 – 5i -12 + 5i • Solving Equations with Complex Numbers Some quadratic equations can be solved by isolating the variable and taking the square root of both sides. ©2012, TESCCC 09/06/12 P) Q) Number 2 51 i 3 5+ i 3 2 3 Conjugate + 51 i 5 −i 3 x 2 = 100 Sample: x = ± 100 x = ± 10i page 3 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Why are the solutions complex conjugates? 6 x 2 + 96 = 0 R) Solve x = 4i or x = -4i ©2012, TESCCC Answers will vary as to explaining why they are complex conjugates. Sample: Have positive and negative imaginary part. 09/06/12 page 4 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Formula KEY Describe the complex numbers below by filling in the table. Chart A 121 1) 2) -56 3) 85 + 169 16 49 4) 5) 5 6) 3 – 14i – 9 + 5i 7) (5 + 2i) 2 Simplified Real Part Imaginary Part Complex Conjugate 11i 0 11i -11i -56 -56 0 -56 85 +13i 85 13i 85 – 13i − 4 i 7 4 i 7 4 i 7 0 − i 5 0 i 5 −i 5 -6 – 9i -6 -9i -6 + 9i 21 + 20i 21 20i 21 – 20i Solve the following quadratic equations and describe the quadratic solutions. 8) 2 x 2 + 50 = 0 9) 5i Complex conjugates 10) 6i Complex conjugates What is the quadratic equation with solutions of (x – (7i)) (x – (-7i)) = 0 (x – 7i) (x + 7i) = 0 x2 + 7ix – 7ix – 49i2 = 0 x2 – 49i2 = 0 x2 + 49 = 0 ©2012, TESCCC 3 x 2 8 = 100 ± 7i 04/18/13 ? page 5 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Formula KEY Part III: Quadratic Formula Any quadratic equation—including those with imaginary or complex solutions—can be solved with the quadratic formula. If ax2 + bx + c = 0, and a 0, then Steps: • Get one side of the equation equal to zero. Quadratic 2 • Identify the values of a, b, and c. b ± b 4ac Formula x= • Substitute these values into the formula. 2a • Simplify to obtain the two solutions. b 2 4ac From the quadratic formula, is called the discriminant. Because this part of the formula lies under the radical, it determines the type of solutions the quadratic equation will have. If… Then the equation has… Sample x + 6x + 9 = 0 a = 1, b = 6, c = 9 Previous Methods x 2 + 6x + 9 = 0 2 b 2 4ac = 0 b 2 4ac > 0 One rational double root , b 2 4ac and is a perfect square (1, 4, 9, 16, 25, etc.) b 2 4ac > 0 , b 2 4ac and is a non-square (2, 5, 8, 12, etc.) b 2 4ac < 0 Two rational roots (6) 4(1)(9 ) = 0 Two complex roots (Conjugates involving imaginary numbers) x = -3 (double root) 2x 2 + x − 3 = 0 2x 2 + x 3 = 0 a = 2, b = 1, c = 3 (2 x + 3)( x − 1) = 0 3 x = − , x =1 2 (1) 2 4(2)( 3) = 25 x 2 10 x + 23 = 0 Two irrational roots ( x + 3)( x + 3) = 0 2 a = 1, b = 10, c = 23 • • Will not factor Graphs can provide approximations only 5± 2 (Here, x = ) • • Will not factor Graphs cannot provide solutions 2 ± 6i (Here, x = ) ( 10) 2 4(1)( 23 ) = 8 x 2 4 x + 40 = 0 a = 1, b = 4, c = 40 ( 4) 2 4(1)( 40) = 144 Example Problems For each problem, a) Write the equation in the form ax2 + bx + c = 0. b) Find the discriminant, and use it to describe the nature of the roots of the equation. c) Complete the quadratic formula to find the solutions. ©2012, TESCCC 04/18/13 page 6 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 1) 3x 2 + 7x = 2 ©2012, TESCCC a) 3x2 + 7x + 2 = 0 b) Discriminant =25 (Two rational roots) 1 − 3 c) Roots: {-2, } 04/18/13 page 7 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Formula KEY 2) x 2 2x = 2x 2 a) x2 – 4x + 2 = 0 b) Discriminant = 8 (Two irrational roots) 4± 8 2 c) Roots: x = , or approximately {3.4, 0.6} 3) x 2 = 14 x 58 a) x2 – 14x + 58 = 0 b) Discriminant = -36 (Two complex conjugate roots) c) Roots: {7 + 3i, 7 – 3i} 4) Given a quadratic equation with rational coefficients and a root of 3 + 5i a) What is another root to the equation? b) What are the factors of the equation? c) What is the quadratic equation with the given root? d) What is the representative function? a) 3 – 5i b) (x – (3 + 5i))(x – (3 – 5i)) c) x2 – 6x + 34 = 0 d) y = x2 – 6x + 34 5) Given a quadratic equation with rational coefficients and a root of -2 + 8i a) What is another root to the equation? b) What are the factors of the equation? c) What is the quadratic equation with the given root? d) What is the representative function? a) -2 – 8i b) (x – (-2 + 8i))(x – (-2 – 8i)) c) x2 + 4x + 68 = 0 ©2012, TESCCC 04/18/13 page 8 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 d) y = x2 + 4x + 68 ©2012, TESCCC 04/18/13 page 9 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Formula KEY Practice Problems For each problem, Use the discriminant to describe the nature of the roots of the equation. Complete the quadratic formula to find the solutions. 1) x 2 − 8 x + 12 = 0 2) Discriminant > 0, perfect square Two rational roots: {6, 2} 3 x 2 + x = 10 Discriminant > 0, perfect square 5 3 Two rational roots: {-2, 3) x 2 + 25 = 10 x 4) Discriminant = 0 One rational double root: {5} 5) x 2 + 6 = 8x 10 x 2 + 2 x + 5 = 0 Discriminant < 0 Two complex conjugate roots: −1 ± 7i 10 { } or {0.1 + 0.7i, 0.1 – 0.7i} 6) Discriminant > 0, non-square 4 ± 10 Two irrational roots: { } 2 x 2 + x − 14 = −11 Discriminant > 0, perfect square 3 2 Two rational roots: {- , 1} 7) Given a quadratic equation with rational coefficients and a root of 2 3i a) What is another root to the equation? ( x − 32 i )( x + 32 i ) b) What are the factors of the equation? x 2 + 94 = 0 c) What is the quadratic equation with the given root? ©2012, TESCCC } 04/18/13 2 − i 3 page 10 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 8) Given a quadratic equation with rational coefficients and a root of 5 – i a) What is another root to the equation? 5 + i b) What are the factors of the equation? (x – (5 – i)) (x – (5 + i)) c) What is the quadratic equation with the given root? x2 – 10x + 26 = 0 ©2012, TESCCC 04/18/13 page 11 of 11