quadratic formula
DESCRIPTION
Quadratic formula. Warm Up Write each function in standard form. Evaluate b 2 – 4 ac for the given values of the valuables. 2. g ( x ) = 2( x + 6) 2 – 11. 1. f ( x ) = ( x – 4) 2 + 3 . g ( x ) = 2 x 2 + 24 x + 61. f ( x ) = x 2 – 8 x + 19 . 4. a = 1, b = 3, c = –3. - PowerPoint PPT PresentationTRANSCRIPT
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Quadratic formula
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Warm UpWrite each function in standard form.
Evaluate b2 – 4ac for the given values of the valuables.
1. f(x) = (x – 4)2 + 3 2. g(x) = 2(x + 6)2 – 11
f(x) = x2 – 8x + 19 g(x) = 2x2 + 24x + 61
4. a = 1, b = 3, c = –33. a = 2, b = 7, c = 59 21
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Solve quadratic equations using the Quadratic Formula.Classify roots using the discriminant.
Objectives
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discriminantVocabulary
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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.
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To subtract fractions, you need a common denominator.
Remember!
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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.
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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.
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Find the zeros of f(x)= 2x2 – 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.
2x2 – 16x + 27 = 0
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Check Solve by completing the square.
Example 1 Continued
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Find the zeros of f(x) = x2 + 3x – 7 using the Quadratic Formula.
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 1 for a, 3 for b, and –7 for c.
Simplify.
Write in simplest form.
Check It Out! Example 1a
x2 + 3x – 7 = 0
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Check Solve by completing the square.x2 + 3x – 7 = 0
x2 + 3x = 7
Check It Out! Example 1a Continued
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Find the zeros of f(x)= x2 – 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
x2 – 8x + 10 = 0
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Check Solve by completing the square.x2 – 8x + 10 = 0
x2 – 8x = –10 x2 – 8x + 16 = –10 + 16
(x + 4)2 = 6
Check It Out! Example 1b Continued
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Find the zeros of f(x) = 4x2 + 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)= 4x2 + 3x + 2
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Find the zeros of g(x) = 3x2 – x + 8 using the Quadratic Formula.
Set f(x) = 0
Write the Quadratic Formula.
Substitute 3 for a, –1 for b, and 8 for c.
Simplify.
Write in terms of i.
Check It Out! Example 2
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The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.
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Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.
Caution!
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Find the type and number of solutions for the equation.
Example 3A: Analyzing Quadratic Equations by Using the Discriminant
x2 + 36 = 12xx2 – 12x + 36 = 0b2 – 4ac(–12)2 – 4(1)(36)144 – 144 = 0
b2 – 4ac = 0
The equation has one distinct real solution.
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Find the type and number of solutions for the equation.
Example 3B: Analyzing Quadratic Equations by Using the Discriminant
x2 + 40 = 12xx2 – 12x + 40 = 0b2 – 4ac(–12)2 – 4(1)(40)144 – 160 = –16
b2 –4ac < 0
The equation has two distinct nonreal complex solutions.
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Find the type and number of solutions for the equation.
Example 3C: Analyzing Quadratic Equations by Using the Discriminant
x2 + 30 = 12xx2 – 12x + 30 = 0b2 – 4ac(–12)2 – 4(1)(30)144 – 120 = 24
b2 – 4ac > 0
The equation has two distinct real solutions.
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Check It Out! Example 3a Find the type and number of solutions for the equation.
x2 – 4x = –4x2 – 4x + 4 = 0b2 – 4ac(–4)2 – 4(1)(4)16 – 16 = 0
b2 – 4ac = 0
The equation has one distinct real solution.
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Check It Out! Example 3b Find the type and number of solutions for the equation.
x2 – 4x = –8x2 – 4x + 8 = 0b2 – 4ac(–4)2 – 4(1)(8)16 – 32 = –16
b2 – 4ac < 0
The equation has two distinct nonreal complex solutions.
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Check It Out! Example 3c Find the type and number of solutions for each equation.
x2 – 4x = 2x2 – 4x – 2 = 0b2 – 4ac(–4)2 – 4(1)(–2)16 + 8 = 24
b2 – 4ac > 0
The equation has two distinct real solutions.
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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.
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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 = –16t2 + 24.6t + 6.5. 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
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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 = –16t2 + 24.6t + 6.5
Example 4 Continued
Set y equal to 0.0 = –16t2 + 24.6t + 6.5
Use the Quadratic Formula.
Substitute –16 for a, 24.6 for b, and 6.5 for c.
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Example 4 Continued
Simplify.
The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.
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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 ≈ 51.86x ≈ 52
The shot put will have traveled a horizontal distance of about 52 feet.
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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 = 1.77.
y = –16t2 + 24.6t + 6.5 y ≈ –16(1.77)2 + 24.6(1.77) + 6.5 y ≈ –50.13 + 43.54 + 6.5 y ≈ –0.09
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Check It Out! Example 4
The pilot’s altitude decreases, which changes the function describing the water’s height toy = –16t2 –2t + 400. To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water?
A pilot of a helicopter plans to release a bucket of water on a forest fire. The height y in feet of the water t seconds after its release is modeled by y = –16t2 – 2t + 500. the horizontal distance x in feet between the water and its point of release is modeled by x = 91t.
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Step 1 Use the equation to determine how long it will take the water to hit the ground. Set the height of the water equal to 0 feet, and then use the quadratic formula for t.
y = –16t2 – 2t + 400Set y equal to 0.0 = –16t2 – 2t + 400
Write the Quadratic Formula.
Substitute –16 for a, –2 for b, and 400 for c.
Check It Out! Example 4 Continued
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Check It Out! Example 4
Simplify.
t ≈ –5.063 or t ≈ 4.937
The time cannot be negative, so the water lands on a target about 4.937 seconds after it is released.
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Check It Out! Example 4 Continued
Simplify.
Step 2 The horizontal distance x in feet between the water and its point of release is modeled by x = 91t. Find the horizontal distance that the water will have traveled in this time.
x = 91tSubstitute 4.937 for t.x ≈ 91(4.937)
x ≈ 449.267x ≈ 449
The water will have traveled a horizontal distance of about 449 feet. Therefore, the pilot should start releasing the water when the horizontal distance between the helicopter and the fire is 449 feet.
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Check It Out! Example 4 Continued Check Use substitution to check that the water hits
the ground after about 4.937 seconds.
The height is approximately equal to 0 when t = 4.937.
y = –16t2 – 2t + 400 y ≈ –16(4.937)2 – 2(4.937) + 400y ≈ –389.983 – 9.874 + 400y ≈ 0.143
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Properties of Solving Quadratic Equations
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Properties of Solving Quadratic Equations
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No matter which method you use to solve a quadratic equation, you should get the same answer.
Helpful Hint
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Lesson Quiz: Part IFind the zeros of each function by using the Quadratic Formula.
1. f(x) = 3x2 – 6x – 5
2. g(x) = 2x2 – 6x + 5
Find the type and member of solutions for each equation.
3. x2 – 14x + 50 4. x2 – 14x + 482 distinct nonreal complex
2 distinct real
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Lesson Quiz: Part II5. A pebble is tossed from the top of a cliff. The
pebble’s height is given by y(t) = –16t2 + 200, where t is the time in seconds. Its horizontal distance in feet from the base of the cliff is given by d(t) = 5t. How far will the pebble be from the base of the cliff when it hits the ground?about 18 ft