5.5 and 5.6 Solving Quadratics with Complex Roots by square root and completing the square method Solving Quadratics using Quadratic Formula

Warm Up- Simplify each expression by hand f(x) = x 2 – 18x + 16 g(x) = 3x 2 – 18x + 7 Find the zeros of each function by graphing, factoring, and completing the square.

Write the function in vertex form, and identify its vertex Factor so the coefficient of x 2 is 1. g(x) = 5(x 2 – 10x) Set up to complete the square. Add. Because is multiplied by 5, you must subtract 5. Check It Out! Example 4b g(x) = 5x 2 – 50x g(x) = 5(x 2 – 10x + ) –

Define and use imaginary and complex numbers. Solve quadratic equations with complex roots. 5.5 Objectives

imaginary unit imaginary number complex number real part imaginary part complex conjugate Vocabulary

You can see in the graph of f(x) = x below that f has no real zeros. If you solve the corresponding equation 0 = x 2 + 1, you find that x =,which has no real solutions. However, you can find solutions if you define the square root of negative numbers, which is why imaginary numbers were invented. The imaginary unit i is defined as. You can use the imaginary unit to write the square root of any negative number.

Express the number in terms of i. Example 1A: Simplifying Square Roots of Negative Numbers Factor out –1. Product Property. Simplify. Multiply. Express in terms of i.

Express the number in terms of i. Example 1B: Simplifying Square Roots of Negative Numbers Factor out –1. Product Property. Simplify. Express in terms of i.

Check It Out! Example 1a Express the number in terms of i. Factor out –1. Product Property. Simplify. Express in terms of i. Product Property.

Check It Out! Example 1b Express the number in terms of i. Factor out –1. Product Property. Simplify. Express in terms of i. Multiply.

Solve the equation. Example 2A: Solving a Quadratic Equation with Imaginary Solutions Take square roots. Express in terms of i. Check x 2 = –144 –144 (12i) 2 144i 2 144(–1) x 2 = –144 – (–1) 144i 2 (–12i) 2

Solve the equation. Example 2B: Solving a Quadratic Equation with Imaginary Solutions Add –90 to both sides. Divide both sides by 5. Take square roots. Express in terms of i. Check 5x = (18)i (–1) +90 5x = 0

Check It Out! Example 2c 9x = 0 Solve the equation. Add –25 to both sides. Divide both sides by 9. Take square roots. Express in terms of i. 9x 2 = –25

Every complex number has a real part a and an imaginary part b. A complex number is a number that can be written in the form a + bi, where a and b are real numbers and i =. The set of real numbers is a subset of the set of complex numbers C.

Real numbers are complex numbers where b = 0. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. These are sometimes called pure imaginary numbers. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Find the values of x and y that make the equation 4x + 10i = 2 – (4y)i true. Example 3: Equating Two Complex Numbers Equate the real parts. 4x + 10i = 2 – (4y)i Real parts Imaginary parts 4x = 2 Solve for y. Solve for x. Equate the imaginary parts. 10 = –4y

Find the zeros of the function. Example 4A: Finding Complex Zeros of Quadratic Functions x x + = –26 + f(x) = x x + 26 Add to both sides. x x + 26 = 0 Set equal to 0. Rewrite. x x + 25 = – (x + 5) 2 = –1 Factor. Take square roots. Simplify.

Find the zeros of the function. Example 4B: Finding Complex Zeros of Quadratic Functions Add to both sides. g(x) = x 2 + 4x + 12 x 2 + 4x + 12 = 0 x 2 + 4x + = –12 + x 2 + 4x + 4 = – Rewrite. Set equal to 0. Take square roots. Simplify. Factor. (x + 2) 2 = –8

If a quadratic equation with real coefficients has nonreal roots, those roots are complex conjugates. When given one complex root, you can always find the other by finding its conjugate. Helpful Hint The solutions and are related. These solutions are a complex conjugate pair. Their real parts are equal and their imaginary parts are opposites. The complex conjugate of any complex number a + bi is the complex number a – bi.

Find each complex conjugate. Example 5: Finding Complex Zeros of Quadratic Functions A i B. 6i 0 + 6i 8 + 5i 8 – 5i Write as a + bi. Find a – bi. Write as a + bi. Find a – bi. Simplify. 0 – 6i –6i

Solve quadratic equations using the Quadratic Formula. Classify roots using the discriminant. 5.8 Objectives

discriminant Vocabulary

You have learned several methods for solving quadratic equations: graphing, making tables, factoring, using square roots, and completing the square. Another method is to use the Quadratic Formula, which allows you to solve a quadratic equation in standard form. By completing the square on the standard form of a quadratic equation, you can determine the Quadratic Formula.

To subtract fractions, you need a common denominator. Remember!

The symmetry of a quadratic function is evident in the last step,. These two zeros are the same distance,, away from the axis of symmetry,,with one zero on either side of the vertex.

You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions.

Find the zeros of f(x)= 2x 2 – 16x + 27 using the Quadratic Formula. Example 1: Quadratic Functions with Real Zeros Set f(x) = 0. Write the Quadratic Formula. Substitute 2 for a, –16 for b, and 27 for c. Simplify. Write in simplest form. 2x 2 – 16x + 27 = 0

Check Solve by completing the square. Example 1 Continued

Find the zeros of f(x)= x 2 – 8x + 10 using the Quadratic Formula. Set f(x) = 0. Write the Quadratic Formula. Substitute 1 for a, –8 for b, and 10 for c. Simplify. Write in simplest form. Check It Out! Example 1b x 2 – 8x + 10 = 0

Check Solve by completing the square. x 2 – 8x + 10 = 0 x 2 – 8x = –10 x 2 – 8x + 16 = – (x + 4) 2 = 6 Check It Out! Example 1b Continued

Find the zeros of f(x) = 4x 2 + 3x + 2 using the Quadratic Formula. Example 2: Quadratic Functions with Complex Zeros Set f(x) = 0. Write the Quadratic Formula. Substitute 4 for a, 3 for b, and 2 for c. Simplify. Write in terms of i. f(x)= 4x 2 + 3x + 2

The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.

Make sure the equation is in standard form before you evaluate the discriminant, b 2 – 4ac. Caution!

Find the type and number of solutions for the equation. Example 3A: Analyzing Quadratic Equations by Using the Discriminant x = 12x x 2 – 12x + 36 = 0 b 2 – 4ac (–12) 2 – 4(1)(36) 144 – 144 = 0 b 2 – 4ac = 0 The equation has one distinct real solution.

Find the type and number of solutions for the equation. Example 3B: Analyzing Quadratic Equations by Using the Discriminant x = 12x x 2 – 12x + 40 = 0 b 2 – 4ac (–12) 2 – 4(1)(40) 144 – 160 = –16 b 2 –4ac < 0 The equation has two distinct nonreal complex solutions.

Find the type and number of solutions for the equation. Example 3C: Analyzing Quadratic Equations by Using the Discriminant x = 12x x 2 – 12x + 30 = 0 b 2 – 4ac (–12) 2 – 4(1)(30) 144 – 120 = 24 b 2 – 4ac > 0 The equation has two distinct real solutions.

The graph shows related functions. Notice that the number of real solutions for the equation can be changed by changing the value of the constant c.

An athlete on a track team throws a shot put. The height y of the shot put in feet t seconds after it is thrown is modeled by y = –16t t The horizontal distance x in between the athlete and the shot put is modeled by x = 29.3t. To the nearest foot, how far does the shot put land from the athlete? Example 4: Sports Application

Step 1 Use the first equation to determine how long it will take the shot put to hit the ground. Set the height of the shot put equal to 0 feet, and the use the quadratic formula to solve for t. y = –16t t Example 4 Continued Set y equal to 0. 0 = –16t t Use the Quadratic Formula. Substitute –16 for a, 24.6 for b, and 6.5 for c.

Example 4 Continued Simplify. The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.

Step 2 Find the horizontal distance that the shot put will have traveled in this time. x = 29.3t Example 4 Continued Substitute 1.77 for t. Simplify. x ≈ 29.3(1.77) x ≈ x ≈ 52 The shot put will have traveled a horizontal distance of about 52 feet.

Check Use substitution to check that the shot put hits the ground after about 1.77 seconds. Example 4 Continued The height is approximately equal to 0 when t = y = –16t t y ≈ –16(1.77) (1.77) y ≈ – y ≈ –0.09

Properties of Solving Quadratic Equations

No matter which method you use to solve a quadratic equation, you should get the same answer. Helpful Hint

Homework Guided practice 5.5 and 5.6 Evens Tomorrow review of homework and additional questions (1/2 block) Quiz tomorrow. Tutoring Today. Also, before tutoring on today we will have a quick review (bring your factoring quiz) and retakes for the quiz.