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SAT Problem of the Day

Learn how to classify and find roots of quadratic equations, and perform operations on complex numbers. Practice problems included.

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SAT Problem of the Day

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  1. SAT Problem of the Day

  2. 5.6 Quadratic Equations and Complex Numbers Objectives: Classify and find all roots of a quadratic equation

  3. Let ax2 + bx + c = 0, where a = 0. Solutions of a Quadratic Equation The expression b2 – 4ac is called the discriminant. If b2 – 4ac > 0, then the quadratic equation has 2 distinct real solutions. If b2 – 4ac = 0, then the quadratic equation has 1 real solution. If b2 – 4ac < 0, then the quadratic equation has 0 real solutions.

  4. Example 1 Find the discriminant for each equation. Then determine the number of real solutions. a) 3x2 – 6x + 4 = 0 b2 – 4ac = (-6)2 – 4(3)(4) = 36 – 48 = -12 no real solutions b) 3x2 – 6x + 3 = 0 b2 – 4ac = (-6)2 – 4(3)(3) = 36 – 36 = 0 one real solution c) 3x2 – 6x + 2 = 0 b2 – 4ac = (-6)2 – 4(3)(2) = 36 – 24 = 12 two real solutions

  5. Practice Identify the number of real solutions: 1) -3x2 – 6x + 15 = 0

  6. The imaginary unit is defined as and i2 = -1. If r > 0, then the imaginary number is defined as follows: Imaginary Numbers

  7. Example 2 Solve 6x2 – 3x + 1 = 0.

  8. Practice Solve -4x2 + 5x – 3 = 0.

  9. Homework Lesson 5.6 exercises 19-35 Odd

  10. SAT Problem of the Day

  11. 5.6 Quadratic Equations and Complex Numbers Objectives: Graph and perform operations on complex numbers

  12. A complex number is any number that can be written as a + bi, where a and b are real numbers and Imaginary Numbers a is called the real part and b is called the imaginary part. 3 + 4i 3 4i imaginary part real part

  13. Example 1 Find x and y such that -3x + 4iy = 21 – 16i. Real parts Imaginary parts -3x = 21 4y = -16 y = -4 x = -7 x = -7 and y = -4

  14. Example 2 Find each sum or difference. a) (-10 – 6i) + (8 – i) = (-10 + 8) + (-6i – i) = -2 – 7i b) (-9 + 2i) – (3 – 4i) = (-9 – 3) + (2i + 4i) = -12 + 6i

  15. Example 3 Multiply. (2 – i)(-3 – 4i) = -6 - 8i + 3i + 4i2 = -6 - 5i + 4(-1) = -10 – 5i

  16. The conjugate of a + bi is denoted a + bi. Conjugate of a Complex Number The conjugate of a complex number a + bi is a – bi.

  17. Example 4 multiply by 1, using the conjugate of the denominator (3 – 2i) (-4 – i) = (-4 + i) (-4 - i) -12 - 3i + 8i + 2i2 = 16 + 4i - 4i - i2 -12 + 5i + 2(-1) -14 + 5i = = 16 - (-1) 17

  18. Practice

  19. Homework Lesson 5.6 Exercises 49-57 odd, 65, 67, 71, 75

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