quadratic functions and equations - ms....
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Quadratic Functions and Equations
Quadratic Graphs and Their Properties
Objective: To graph quadratic functions of the form y = ax2 and y = ax2 + c.
Objectives
β’ I can identify a vertex.
β’ I can grapy π¦ = ππ₯2.
β’ I can compare widths of parabolas.
β’ I can graph π¦ = ππ₯2 + π.
β’ I can use the falling object model.
Vocabularyβ’ A quadratic function is a type of nonlinear function that models certain
situations where the rate of change is not constant.
β’ The graph of a quadratic function is a symmetric curve with the highest or lowest point corresponding to a maximum or minimum value.
β’ Standard Form of a Quadratic Function:β’ A quadratic function is a function that can be written in the form π¦ = ππ₯2 + ππ₯ + π,
where a β 0. This form is called the standard form of a quadratic function.
β’ Examples: π¦ = 3π₯2 π¦ = π₯2 + 9 π¦ = π₯2 β π₯ β 2
Vocabulary
β’ The simplest quadratic function π π₯ = π₯2 ππ π¦ = π₯2 is the quadratic parent function.
β’ The graph of a quadratic function is a U-shaped curve called a parabola.
β’ You can fold a parabola so that the two sides match exactly. This property is called symmetry.
β’ The fold or line that divides the parabola into two matching halves is called the axis of symmetry.
Vocabulary
β’ The highest or lowest point of a parabola is its vertex, what is on the axis of symmetry.
β’ If a > 0 in y = ax2 + bx + c, the parabola opens upward. The vertex is the minimum point, or lowest point, of the parabola.
β’ If a < 0 in y = ax2 + bx + c , the parabola opens downward. The vertex is the maximum point, or highest point, of the parabola.
Identifying a Vertex
PracticeIdentify the vertex.
Practice
Vocabulary
β’ You can use the fact that a parabola is symmetric to graph it quickly.
β’ First, find the coordinates of the vertex and several points on one side of the vertex.
β’ Then reflect the points across the axis of symmetry.
β’ For graphs of functions of the form y = ππ₯2, the vertex is at the origin.
β’ The axis of symmetry is the yβaxis, or x = 0.
Graphing π¦ = ππ₯2
Graph the function. Make a table of values. What are the domain and range?
1. π¦ =1
3π₯2
2. π¦ = β3π₯2
3. π¦ = 4π₯2
Practice
Graph each function. Then identify the domain and range of the function.
1. π¦ = β4π₯2
2. π¦ = β1
3π₯2
3. π π₯ = 1.5π₯2
4. π π₯ =2
3π₯2
Vocabulary
β’ The coefficient of the x2βterm in a quadratic function affects the width of the parabola as well as the direction in which it opens.
β’ When π < π , the graph of π¦ = ππ₯2 is wider than the graph of π¦ = ππ₯2.
Comparing Widths of Parabolas
Use your calculator to graph. What is the order, from widest to narrowest, of the graphs of the quadratic functions.
π π₯ = β4π₯2, π π₯ =1
4π₯2, π(π₯) = π₯2
π π₯ = βπ₯2, π π₯ = 3π₯2, π π₯ = β1
3π₯2
Practice
Order each group of quadratic functions from widest to narrowest graph.β’ π¦ = 3π₯2, π¦ = 2π₯2, π¦ = 4π₯2
β’ π¦ = β1
2π₯2, π¦ = 5π₯2, π¦ = β
1
4π₯2
β’ π π₯ = 5π₯2, π π₯ = β3π₯2, π π₯ = π₯2
β’ π π₯ = β2π₯2, π π₯ = β2
3π₯2, π π₯ = β4π₯2
Vocabulary
β’ The y β axis is the axis of symmetry for graphs of functions of the form π¦ = ππ₯2 + π.
β’ The value of c translates the graph up or down.
Graphing π¦ = ππ₯2 + π
What is the relationship of the following graphs?1. π¦ = 2π₯2 + 3 πππ π¦ = 2π₯2
2. π¦ = π₯2 πππ π¦ = π₯2 β 3
3. π¦ = β1
2π₯2 πππ π¦ = β
1
2π₯2 + 1
Practice
Graph each function.1. π π₯ = π₯2 + 4
2. π π₯ = βπ₯2 β 3
3. π π₯ =1
2π₯2 + 2
Vocabulary
As an object falls, its speed continues to increase, so its height above the ground decreases at a faster and faster rate.
Ignoring air resistance, you can model the objectβs height with the function h = β16t2 + c.
The height h is in feet, the time t is in seconds, and the objectβs initial height c is in feet.
Using the Falling Object Model
An acorn drops from a tree branch 20 feet above the ground. The function h = β16t2 + 20 gives the height h of the acorn (in feet) after t seconds. What is the graph of this quadratic function? At what time does the the acorn hit the ground? t h = β16t2 + 20
Using the Falling Object Model
An acorn drops from a tree branch 70 feet above the ground. The function h = β16t2 + 70 gives the height h of the acorn (in feet) after t seconds. What is the graph of this quadratic function? At what time does the the acorn hit the ground? t h = β16t2 + 70
Practice
A person walking across a bridge accidentally drops an orange into the river below from a height of 40 feet. The function β = β16π‘2 + 40 gives the orangeβs approximate height h above the water, in feet, after t seconds. In how many seconds will the orange hit the water?
A bird drops a stick to the ground from a height of 80 feet. The function β =β16π‘2 + 80 gives the stickβs approximate height h above the ground, in feet, after t seconds. Graph the function. At about what time does the stick hit the ground?
Quadratic Functions
Objective: To graph quadratic functions of the form y = ax2 + bx + c.
Objectives
β’ I can graph y = ππ₯2 + ππ₯ + π.
β’ I can use the vertical motion model.
Vocabularyβ’ In the quadratic function y = ax2 + bx + c, the value of b affects the position of the axis of
symmetry.
β’ The axis of symmetry changes with each equation because of the change in the b-value. The equation of the axis of symmetry is related to the ratio
π
π.
β’ The equation of the axis of symmetry is π₯ = β1
2
π
πππ π₯ =
βπ
2π.
β’ Graph of a Quadratic Function
o The graph of y = ax2 + bx + c, where a β 0, has the line π₯ =βπ
2πas its axis of symmetry.
The xβcoordinate of the vertex is βπ
2π.
Vocabulary
β’ When you substitute x = 0 into the equation y = ax2 + bx + c, you get y = c. So the yβintercept of a quadratic function is c.
β’ You can use the axis of symmetry and the yβintercept to help you graph a quadratic function.
Graphing y = ax2 + bx + c
What is the graph of the function? Show the axis of symmetry.1. π¦ = π₯2 β 6π₯ + 4
2. π¦ = βπ₯2 + 4π₯ β 2
3. π¦ = 2π₯2 + 3
4. π¦ = β3π₯2 + 12π₯ + 1
5. π π₯ = π₯2 + 4π₯ β 5
6. π π₯ = β4π₯2 + 11
Practice
What is the graph of the function? Show the line of symmetry. 1. π¦ = 2π₯2 β 6π₯ + 1
2. π π₯ = 2π₯2 + 4π₯ β 1
3. π¦ = 6π₯2 + 6π₯ β 5
4. π π₯ = β5π₯2 + 3π₯ + 2
5. π¦ = β2π₯2 β 10π₯
6. π¦ = β4π₯2 β 16π₯ β 3
Vocabulary
β’ You have used h = β16t2 + c to find the height h above the ground of an object falling from an initial height c at time t.
β’ If an object projected into the air given an initial upward velocity vcontinues with no additional force of its own, the formula h = β16t2 + vt + c givens its approximate height above the ground.
Using a Vertical Motion Model
During halftime of a basketball game, a sling shot launches Tβshirts at the crowd. A Tβshirt launched with an initial upward velocity of 72 feet per second. The Tβshirt is caught 35 feet above the court. The Tβshirt is launched from a height of 5 feet.
a. How long will it take the Tβshirt to reach its maximum height?
b. What is the maximum height?
c. What is the range of the function that models the height of the Tβshirt over time?
Using a Vertical Motion Model
During halftime of a basketball game, a sling shot launches Tβshirts at the crowd. A Tβshirt launched with an initial upward velocity of 64 feet per second. The Tβshirt is caught 35 feet above the court. The Tβshirt is launched from a height of 5 feet.
a. How long will it take the Tβshirt to reach its maximum height?
b. What is the maximum height?
c. What is the range of the function that models the height of the Tβshirt over time?
Practice
A baseball is thrown into the air with an upward velocity of 30 feet per second. Its height h, in feet, after t seconds is given by the function h = β16t2 + 30t + 6.
a. How long will it take the ball to reach its maximum height?
b. What is the ballβs maximum height?
c. What is the range of the function?
Solving Quadratic Equations
Objective: To solve quadratic equations by graphing and using square roots.
Objectives
β’ I can solve by graphing.
β’ I can solve using square roots.
β’ I can choose a reasonable solution.
Vocabularyβ’ Standard Form of a Quadratic Equation:
A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where a β 0. This form is called the standard form of a quadratic equation.
β’ Quadratic equations can be solved by a variety of methods, including graphing and finding square roots.
β’ One way to solve a quadratic equation ax2 + bx + c = 0 is to graph the related quadratic function y = ax2 +bx + c. The solutions of the equation are the x β intercepts of the related function.
Vocabulary
β’ A quadratic equation can have two, one, or no real-number solutions.
β’ The solutions of a quadratic equation and the xβintercepts of the graph of the related function are often called roots of the equation or zeros of the function.
Solving by Graphing
What are the solutions of each equation? Use a graph of the related function.
1. x2 β 1 = 0
2. x2 = 0
3. x2 + 1 = 0
4. x2 β 16 = 0
5. 3x2 + 6 = 0
6. x2 β 25 = β25
PracticeSolve each equation by graphing the related function. If the equation has no realβnumber solution, write no solution.
1. π₯2 β 9 = 0
2. 3π₯2 = 0
3.1
3π₯2 β 3 = 0
4. π₯2 + 5 = 5
5. π₯2 β 10 = β10
6. 2π₯2 β 18 = 0
Vocabulary
β’ You can solve equations of the form x2 = k by finding the square roots of each side.
Solving Using Square Roots
What are the solutions?
1. 3x2 β 75 = 0
2. m2 β 36 = 0
3. 3x2 + 15 = 0
4. 4d2 + 16 = 16
5. t2 = 144
6. y2 β 225 = 0
7. x2 β 25 = 0
8. 2x2 β 8 = 0
Practice
Solve each equation by finding square roots. If the question has no realβnumber solution, write no solution.
1. π2 = 81
2. π€2 β 36 = β64
3. 64π2 = 16
4. 5π2 β 20 = 0
5. 144 β π2 = 0
6. 3π2 + 12 = 0
7. π2 + 49 = 49
8. π2 β 196 = 0
Vocabulary
β’ You can solve some quadratic equations that model real β world problems by finding square roots.
β’ In many cases, the negative square root may not be a reasonable solution.
Choosing a Reasonable Solution
1. An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length l that is twice the width w. The volume of the tank is 420 ft3. What is the width of the tank to the nearest tenth of a foot? (π = ππ€β), h = 3 ft
2. Suppose that the tank has a height of 4 feet and a volume of 500 ft3. What is the width of the tank to the nearest tenth of a foot.
Practice
β’ You have enough paint to cover an area of 50 ftΒ². What is the side length of the largest square that you could paint? Round your answer to the nearest tenth of a foot.
β’ Find the length of a square with an area of 75 ftΒ². Round to your answer to the nearest tenth of a foot.
Factoring to Solve Quadratic Equations
Objective: To solve quadratic equations by factoring.
Objectives
β’ I can use the zero-product property.
β’ I can solve by factoring.
β’ I can write in standard form first.
β’ I can use factoring to solve real-world problems.
Vocabulary
β’ In the previous lesson, you solved quadratic equations ax2 + bx + c = 0 by finding square roots. This method works if b = 0.
β’ You can solve some quadratic equations, including equations where b β 0, by using the ZeroβProduct Property.
β’ The Multiplication Property of Zero states that for any real number a, a Γ 0 = 0. This is equivalent to the following statement: For any real numbers a and b, if a = 0 or b = 0, then ab = 0.
β’ The ZeroβProduct Property reverses this statement.
Vocabulary
ZeroβProduct Property
β’ For any real numbers a and b, if ab = 0, then a = 0 or b = 0.
β’ Example: If (x + 3)(x + 2) = 0, then x + 3 = 0 or x + 2 = 0.
Using the ZeroβProduct Property
What are the solutions of the equation?
1. (4t + 1)(t β 2) = 0
2. (x + 1)(x β 5) = 0
3. (2x + 3)(x β 4) = 0
4. (2y + 1)(y + 14) = 0
5. (7n β 2)(5n β 4) = 0
6. (v β 4)(v β 7) = 0
Practice
Use the Zero β Product Property to solve each equation.1. (x β 9)(x β 8) = 0
2. (4k + 5)(k + 7) = 0
3. n(n + 2) = 0
4. β3n(2n β 5) = 0
5. (7x + 2)(5x β 4) = 0
6. (4a β 7)(3a + 8) = 0
Vocabulary
β’ You can also use the ZeroβProduct Property to solve equations of the form ax2 + bx + c = 0 if the quadratic expression ax2 + bx + c can be factored.
Solving by Factoring
What is the solutions of the equations?
1. x2 + 8x + 15 = 0
2. m2 β 5m β 14 = 0
3. p2 + p β 20 = 0
4. 2a2 β 15a + 18 = 0
5. t2 + 3t β 54 = 0
6. 3y2 β 17y + 24 = 0
Practice
Solve by factoring.1. π₯2 + 11π₯ + 10 = 0
2. π2 + 4π β 32 = 0
3. 3π2 + π β 14 = 0
4. π2 β 4π = 21
5. 2π€2 β 11π€ = β12
6. 16π2 = 81
Vocabulary
β’ Before solving a quadratic equation, you may need to add or subtract terms from each side in order to write the equation in standard form.
β’ Then factor the quadratic expression.
Writing in Standard Form First
What are the solutions?
1. 4x2 β 21x = 18
2. x2 + 14x = β49
3. p2 β 4p = 21
4. 2w2 β 11w = β12
5. 3h2 + 17h = β10
6. 9b2 = 16
Practice
β’ Solve by factoring.1. π₯2 + 13π₯ = β42
2. π2 = 5π
3. π‘2 = β3π‘ + 54
4. 3π¦2 β 17π¦ = β24
5. 7π2 + 16π + 15 = 2π2 + 3
6. 4π2 + 3π = 3π2 β 4π + 18
Using Factoring to Solve a RealβWorld Problem
You are constructing a frame for a rectangular photo (17 in by 11 in).You want the frame to be the same width all the way around and the total area of the frame and photo to be 315 in2. What should the outer dimensions of the frame be?
Using Factoring to Solve a RealβWorld Problem
You are constructing a frame for a rectangular photo (17 in by 11 in).You want the frame to be the same width all the way around and the total area of the frame and photo to be 391 in2. What should the outer dimensions of the frame be?
Practice
β’ A box shaped like a rectangular prism has a volume of 280 inΒ³. Its dimensions are 4 in by (n + 2) in by (n + 5) in. Find n.
β’ You are building a rectangular deck. The area of the deck should be 250 ft2. You want the length of the deck to be 5 feet longer than twice its width. What should the dimensions of the deck be?
Completing the Square
Objective: To solve quadratic equations by completing the square.
Objective
β’ I can find c to complete the square.
β’ I can solve π₯2 + ππ₯ + π = 0.
β’ I can find the vertex by completing the square.
β’ I can complete the square when a β 1.
Vocabulary
β’ You can solve any quadratic equation by first writing it in the form m2 = n.
β’ In general, you can change the expression x2 + bx into a perfectβsquare trinomial
by adding (π
2)2 to x2 + bx. This process is called completing the square.
β’ The process is the same whether b is positive or negative.
Finding c to Complete the Square
What is the value of c so that it is a perfectβsquare trinomial?
1. x2 β 16x + c
2. x2 + 20x + c
3. g2 + 17g + c
4. q2 β 4q + c
5. k2 β 5k + c
6. x2 + 18x + c
Practice
Find the value of c such that each expression is a perfectβsquare trinomial.1. π₯2 + 4π₯ + π
2. π2 β 7a + c
3. π2 + 12π + π
4. π€2 + 18π€ + π
5. π2 β 20π + π
6. π2 β 9π + π
Vocabulary
β’ To solve an equation in the form x2 + bx + c = 0, first subtract the constant term c from each side of the equation.
Solving π₯2 + ππ₯ + π = 0
What is the solution of the equation?1. x2 β 14x + 16 = 0
2. x2 + 9x + 15 = 0
3. m2 + 7m β 294 = 0
4. m2 + 16m = β59
5. g2 + 7g = 144
6. z2 β 2z = 323
Practice
What are the solutions for the following?
1. x2 + 6x = 216
2. t2 β 6t = 247
3. r2 β 4r = 30
4. p2 + 5p β 7 = 0
5. m2 + 12m + 19 = 0
6. w2 β 14w + 13 = 0
Vocabulary
β’ The equation π¦ = (π₯ β β)2+π represents a parabola with vertex (h, k).
β’ You can use the method of completing the square to find the vertex of quadratic functions of the form π¦ = π₯2 + ππ₯ + π.
Finding the Vertex by Completing the Square
Find the vertex by completing the square.1. π¦ = π₯2 + 6π₯ + 8
2. π¦ = π₯2 + 4π₯ + 10
3. π¦ = π₯2 + 12π₯ + 34
4. π¦ = π₯2 + 18π₯ β 307
5. π¦ = π₯2 + 12π₯ β 468
Practice
Find the vertex of each parabola by completing the square.1. π¦ = π₯2 + 4π₯ β 16
2. π¦ = π₯2 + 6π₯ β 7
3. π¦ = π₯2 + 2π₯ β 28
4. π¦ = π₯2 β 2π₯ β 323
Vocabulary
β’ The method of completing the square works when a = 1 in ax2 + bx + c = 0.
β’ To solve an equation when a β 1, divide each side by a before completing the square.
Completing the Square When a β 1
You are planning a flower garden consisting of three square plots surrounded by a 1βfoot border. The total area of the garden and the border is 100 ft2. What is the side length x of each square?
x
x
1 1
1
1
xx
Completing the Square When a β 1
You are planning a flower garden consisting of three square plots surrounded by a 1βfoot border. The total area of the garden and the border is 150 ft2. What is the side length x of each square? Round to the nearest hundredth.
1
1
x x x
x
Practice
Solve each equation by completing the square. If necessary, round to the nearest hundredth.
1. 4a2 β 8a = 24
2. 2y2 β 8y β 10 = 0
3. 5n2 β 3n β 15 = 0
4. 4w2 + 12w β 44 = 0
5. 3r2 + 18r = 21
6. 2v2 β 10v β 20 = 8
The Quadratic Formula and the Discriminant
Objective: To solve quadratic equations using the quadratic formula. To find the number of solutions of a quadratic equation.
Objectives
β’ I can use the quadratic formula.
β’ I can find approximate solutions.
β’ I can choose an appropriate method.
β’ I can use the discriminant.
Vocabulary
β’ You can find the solution(s) of any quadratic equation using the quadratic formula.
β’ Quadratic Formula:
β’ Algebra: If ax2 + bx + c = 0, and a β 0, then π₯ =βπΒ± π2β4ππ
2π.
β’ Example: Suppose 2π₯2 + 3π₯ β 5 = 0. Then a = 2, b = 3, and c = β5. Therefore
π₯ =β(3)Β± (3)2β4(2)(β5)
2(2)
β’ Be sure to write a quadratic in standard form before using the quadratic formula.
β’ Hereβs Why It Works: If you complete the square for the general equation ax2 + bx + c = 0, you can derive the quadratic formula.
β’ Step 1: Write ax2 + bx + c = 0 so that the coefficient of x2 is 1.
β’ ππ₯2 + ππ₯ + π = 0
β’ π₯2 +π
ππ₯ +
π
π= 0 Divide each side by a.
β’ Step 2: Complete the square.
β’ π₯2 +π
ππ₯ = β
π
πSubtract
π
πfrom each side.
β’ π₯2 +π
ππ₯ + (
π
2π)2= β
π
π+ (
π
2π)2 Add (
π
2π)2 to each side.
β’ (π₯ +π
2π)2= β
π
π+
π2
4π2 Write the left side as a square.
β’ (π₯ +π
2π)2= β
4ππ
4π2 +π2
4π2 Multiply βπ
πby
4π
4πto get like denominators.
β’ (π₯ +π
2π)2=
π2β4ππ
4π2 Simplify the right side.
β’ Step 3: Solve the equation for x.
β’ (π₯ +π
2π)2=Β±
π2β4ππ
4π2 Take square roots of each side.
β’ π₯ +π
2π= Β±
π2β4ππ
2πSimplify the right side.
β’ π₯ = βπ
2πΒ±
π2β4ππ
2πSubtract
π
2πfrom each side.
β’ π₯ =βπΒ± π2β4ππ
2πSimplify
Using the Quadratic Formula
What are the solutions using the quadratic formula?1. x2 β 8 = 2x
2. x2 β 4x = 21
3. 2x2 + 5x + 3 = 0
4. 3x2 + 19x = 154
5. 18x2 β 45x β 50 = 0
Practice
Use the quadratic formula to solve each equation.1. 3π₯2 β 41π₯ = β110
2. 5π₯2 + 16π₯ β 84 = 0
3. 2π₯2 β π₯ β 120 = 0
4. 3π₯2 + 44π₯ = β96
5. 5π₯2 β 47π₯ = 156
Vocabulary
β’ When the radicand in the quadratic formula is not a perfect square, you can use a calculator to approximate the solutions of an equation.
Finding Approximate Solutions
1. In the shot put, an athlete throws a heavy metal ball through the air. The arc of the ball can be modeled by the equation y = β0.04x2 + 0.84x + 2, where x is the horizontal distance, in meters, from the athlete and y is the height, in meters, of the ball. How far from the athlete will the ball land?
2. A batter strikes a baseball. The equation y = β0.005x2 + 0.7x + 3.5 models its path, where x is the horizontal distance, in feet, the ball travels and y is the height, in feet, of the ball. How far from the batter will the ball land? Round to the nearest tenth of a foot.
Practice
Use the quadratic formula to solve each equation. Round your answer to the nearest hundredth.
1. π₯2 + 8π₯ + 11 = 0
2. 5π₯2 + 12π₯ β 2 = 0
3. 2π₯2 β 16π₯ = β25
4. 6π₯2 + 9π₯ = 32
Vocabulary
There are many methods for solving a quadratic equation.
Method When to Use
Graphing Use if you have a graphing calculator handy
Square Roots Use if the equation has no x-terms
Factoring Use if you can factor the equation easily
Completing the Square Use if the coefficient of x2 is 1, but you cannot easily factor the equation
Quadratic Formula Use if the equation cannot be factored easily or at all
Choosing an Appropriate Method
Which method(s) would you choose to solve each equation? Explain your reasoning.
1. 3x2 β 9 = 0
2. x2 β x β 30 = 0
3. 6x2 + 13x β 17 = 0
4. x2 β 5x + 3 = 0
5. β16x2 β 50x + 21 = 0
6. x2 β 8x + 12 = 0
7. 169x2 = 36
8. 5x2 + 13x β 1 = 0
Practice
Which method(s) would you choose to solve each equation? Justify your reasoning.
1. π₯2 + 4π₯ β 15 = 0
2. 9π₯2 β 49 = 0
3. 4π₯2 β 41π₯ = 73
4. 3π₯2 β 7π₯ + 3 = 0
5. π₯2 + 4π₯ β 60 = 0
6. β4π₯2 + 8π₯ + 1 = 0
Vocabulary
β’ Quadratic equations can have two, one, or no real-number solutions.
β’ Before you solve a quadratic equations, you can determine how many real-number solutions it has by using the discriminant.
β’ The discriminant is the expression under the radical sign in the quadratic
formula. The discriminant is π2 β 4ππ of π₯ =βπΒ± π2β4ππ
2π
β’ The discriminant of a quadratic equation can be positive, negative, or zero.
Using the Discriminant
Discriminant b2 β 4ac > 0 b2 β 4ac = 0 b2 β 4ac < 0
Example x2 β 6x + 7 = 0 The discriminant (-6)2 β 4(1)(7) = 8 which is positive
x2 β 6x + 9 = 0 The discriminant (-6)2 β 4(1)(9) = 0
x2 β 6x + 11 = 0The discriminant (-6)2 β 4(1)(11) = β8
which is negative
Number of Solutions
There are two real-numbersolutions
There is one real-numbersolutions
There are no real-number solutions
The Discriminant
Using the Discriminant
How many real-number solutions does each have?1. 2x2 β 3x = β5
2. 6x2 β 5x = 7
3. x2 + 3x + 11 = 0
4. 9x2 + 12x + 4 = 0
5. x2 β 15 = 0
Practice
Find the number of realβnumber solutions of each equation.1. π₯2 β 2π₯ + 3 = 0
2. π₯2 + 7π₯ β 5 = 0
3. π₯2 + 2π₯ = 0
4. 3π2 + 4π = 10