Quadratic Equations and Functions

Quadratic Equations and Functions
Chapter 8 Quadratic Equations and Functions

Solving Quadratic Equations by Completing the Square
§ 8.1 Solving Quadratic Equations by Completing the Square

Square Root Property We previously have used factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a2 = b, then

Square Root Property Example: Solve x2 = 49 Solve 2x2 = 4 x2 = 2
Solve (y – 3)2 = 4 y = 3  2 y = 1 or 5

Square Root Property Example: Solve x2 + 4 = 0 x2 = 4
There is no real solution because the square root of 4 is not a real number.

Square Root Property Example: Solve (x + 2)2 = 25 x = 2 ± 5
x = 2 + 5 or x = 2 – 5 x = 3 or x = 7

Square Root Property Example: Solve (3x – 17)2 = 28 3x – 17 =

Completing the Square In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).

Completing the Square Example: x2 – 10x x2 + 16x x2 – 7x
What constant term should be added to the following expressions to create a perfect square trinomial? x2 – 10x add 52 = 25 x2 + 16x add 82 = 64 x2 – 7x add

Completing the Square Example:
We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.

Completing the Square Solving a Quadratic Equation in x by Completing the Square If the coefficient of x2 is 1, go to Step 2. Otherwise, divide both sides of the equation by the coefficient of x2. Isolate all variable terms on one side of the equation. Complete the square for the resulting binomial by adding the square of half of the coefficient of the x to both sides of the equation. Factor the resulting perfect square trinomial and write it as the square of a binomial. Use the square root property to solve for x.

Solving Equations Example: Solve by completing the square.
y2 + 6y = 8 y2 + 6y + 9 = 8 + 9 (y + 3)2 = 1 y + 3 = ± = ± 1 y = 3 ± 1 y = 4 or 2

Solving Equations Example: (y + ½)2 = Solve by completing the square.
y2 + y – 7 = 0 y2 + y = 7 y2 + y + ¼ = 7 + ¼ (y + ½)2 =

Solving Equations Example: Solve by completing the square.
2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½ x2 + 7x = ½ = (x + )2 =

Solving Quadratic Equations by Using the Quadratic Formula
§ 8.2 Solving Quadratic Equations by Using the Quadratic Formula

The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.

The Quadratic Formula Quadratic Formula
A quadratic equation written in the form, ax2 + bx + c = 0, has the solutions

The Quadratic Formula Example:
Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1

Solve x2 + x – = 0 by the quadratic formula. x2 + 8x – 20 = (multiply both sides by 8) a = 1, b = 8, c = -20

Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.

The Discriminant The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

The Discriminant Example:
Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions.

Solving Equations by Using Quadratic Methods
§ 8.3 Solving Equations by Using Quadratic Methods

Solving Quadratic Equations
Solving a Quadratic Equation If the equation is in the form (ax+b)2 = c, use the square root property and solve. If not, go to Step 2. Write the equation in standard form: ax2 + bx + c = 0 Try to solve the equation by the factoring method. If not possible, go to Step 4. Solve the equation by the quadratic formula.

Note that the directions on the previous slide did NOT include completing the square. Completing the square often involves more complicated computations with fractions, which can be avoided by using the quadratic formula.

Example: Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1) Let a = 1, b = –3, c = 1

Example: Solve the following quadratic equation. (multiply both sides by 8)

The steps we detailed in solving quadratic equations will only work if the equation is in an obviously recognizable form. Sometimes, we may have to alter the form of an equation to get it into quadratic form. This might involve substitution into the equation, squaring both sides of the equation, or converting a rational equation.

Example: Solve the following equation. We need to isolate the radical and square both sides of the equation.

Example continued: We now have the equation in a standard quadratic form, so we can use quadratic techniques to solve it. Try factoring the quadratic equation and solve the equation. Now we need to set each of the factors equal to 0 and solve the equations. x – 8 = 0 or x – 2 = 0 which gives us x = 8 or x = 2.

Example continued: Now we substitute the values for x into the original equation. true false So the solution is x = 8.

Example: Solve the following equation.

Example continued: We now have the equation in a standard quadratic form, so we can use quadratic techniques to solve it. Let a = 1, b = -9, c = -6 Now we substitute the values for x into the original equation.

Example continued: true

Example continued: Since both values give true statements when substituted into the original equation, both values are solutions. So the solution is

Example: Solve the following equation. 9x4 + 5x2 – 4 = 0 Substitute w = x2 into the equation. 9w2 + 5w – 4 = 0 Now factor the equation. (9w – 4)(w + 1) = 0 Substitute the original variable back into the equation. (9x2 – 4)(x2 + 1) = 0 = (3x – 2)(3x + 2)(x2 + 1)

Example continued: Now we need to set each of the preceding factors equal to 0 and solve the equations. 3x – 2 = 0 or 3x + 2 = 0 or x2 + 1 = 0 Solving, we get x = ± and x = ± i Now we substitute the values for x into the original equation.

Example continued: true true 9i4 + 5i2 – 4 = 0 9(1) + 5(-1) – 4 = 0 9(-i)4 + 5(-i)2 – 4 = 0 9(1) + 5(-1) – 4 = 0 true true So the solutions are x = ± or x = ± i.

Example: Solve the following equation. 3x2/3 + 11x1/3 = 4 Substitute w = x1/3 into the equation. 3w2 + 11w – 4 = 0 Now factor the equation. (3w – 1)(w + 4) = 0 Substitute the original variable back into the equation. (3x1/3 – 1)(x1/3 + 4) = 0

Example continued: Now we need to set each of the preceding factors equal to 0 and solve the equations. 3x1/3 – 1 = 0 or x1/3 + 4 = 0 Solving for the first equation, we get 3x1/3 = 1 x1/3 = (x1/3)3 = ( )3

Example continued: Solving for the second equation, we get x1/3 = -4 (x1/3)3 = (-4)3 = -64 Now we substitute the values for x into the original equation. 3(-64)2/3 + 11(-64)1/3 = 4 3(16) + 11(-4) = 4 true true So the solutions are x = or x = -64.

Nonlinear Inequalities in One Variable
§ 8.4 Nonlinear Inequalities in One Variable

Quadratic Inequalities
We previously have solved linear inequalities. A quadratic inequality is an inequality that can be written so that one side is a quadratic expression and the other side is 0. Examples 3x2 – 4 > 0 –2x2 – 5x + 7  0 x2 + 4x – 6  x2 – 3 < 0 A solution of a quadratic inequality is a value of the variable that makes the inequality a true statement.

If we attempt to solve a quadratic inequality, such as 3x2 + 5x + 2 < 0 we are looking for values of x that will make this a true statement. If we graph the quadratic equation y = 3x2 + 5x + 2, the points of the parabola that lie below the x-axis would provide values of x where the y-value < 0. Hence, those values of x would satisfy 3x2 + 5x + 2 < 0. Similarly, we could also use the graph to find the values of x that satisfy the inequality 3x2 + 5x + 2 > 0 (the x-values of all points above the x-axis).

The points on the graph above and below the x-axis are separated by points actually on the x-axis. These points would have values of x such that 3x2 + 5x + 2 = 0. However, graphing a quadratic equation could be time consuming if you don’t have a computer or graphing calculator. Fortunately, it is not necessary to graph a quadratic inequality to solve this type of problem. We can construct a number line representing the x-axis and find the region(s) on the number line where the inequality is true.

Solving Quadratic Inequalities
Example: Solve the quadratic inequality 3x2 + 5x + 2 < 0. First we need to know when 3x2 + 5x + 2 = 0. Factor the quadratic equation. (3x + 2)(x + 1) = 0 So x = -2/3 or x = -1. We can then examine the three regions of the number line that are created by these two values.

Example continued: -1 -2 -3 -4 1 A B C All the values of x within the same region will produce quadratic values of the same sign (positive or negative), since the sign of the values on the parabola cannot change without passing through a zero value. So we only need to test a single value within a region to find out the sign of ALL quadratic values within that region.

Example continued: Region Test Point Value Substitute into factored Inequality Truth of Inequality A (-4)(-1) < 0 false B /6 (-1/2)(1/6) < 0 true C (2)(1) < 0 false The values in region B satisfy the inequality, but we cannot include the endpoints, since they will produce values that make the inequality = 0, rather than simply < 0. So the solution is the set (-1, -2/3).

Solving Polynomial Inequalities
Solving a Polynomial Inequality Write the inequality in standard form, then solve the related equation. Separate the number line into regions with the solutions from Step 1. For each region, choose a test point and determine whether its value satisfies the original inequality. The solution set includes all the regions whose test point value is a solution of the inequality. If the inequality symbol is  or , the values from Step 1 are solutions; If the inequality symbol is < or >, the values from Step 1 are not included.

Example: Solve the polynomial inequality x(x – 6)(x + 2)  0. First we need to know when x(x – 6)(x + 2) = 0. Since the polynomial is already factored, we set each factor = 0 and solve. x = 0 or x – 6 = 0 or x + 2 = 0 So x = 0, x = 6 or x = -2. We can then examine the four regions of the number line that are created by these three values.

Example continued: -2 -4 -6 -8 6 A B C D 2 8 4 Region Test Point Value Substitute into factored Inequality Truth of Inequality A (-4)(-10)(-2)  0 false B (-1)(-7)(1)  0 true C (2)(-4)(4)  0 false D (8)(2)(10)  0 true

Example continued: The values in regions B and D satisfy the inequality, but we must include the endpoints, since they will produce values that make the inequality = 0, rather than simply > 0. So the solution is the set [-2, 0]  [6, )

Solving Rational Inequalities
These techniques can be extended to inequalities that contain rational expressions, as well as simply polynomial expressions. In this case, however, not only do you have to find the values that solve the equation when = is substituted for the inequality sign, but also values of x that must be excluded from the solution set because they make the denominator = 0. We have to remember to exclude values in the solution set that will give us a 0 denominator, regardless of the type of inequality sign used.

Solving a Rational Inequality Solve for values that make all denominators 0. Solve the related equation. Separate the number line into regions with the solutions from Steps 1 and 2. For each region, choose a test point and determine whether its value satisfies the original inequality. The solution set includes the regions whose test point value is a solution. Check whether to include values from Step 2. Be sure not to include values that make any denominator 0.

Example: Solve the rational inequality First we need to know when the rational expression is undefined (denominator = 0). x – 10 = 0  x = 10 Then we need to solve the equality

Example continued: The equality will hold when x + 10 =  x = -10. We can then examine the three regions of the number line that are created by these two values. 12 -12 -8 -4 4 A B C 8 16 -16

Example continued: Region Test Point Value Substitute into factored Inequality Truth of Inequality A (-1)/(-21)  0 true B (10)/(-10)  0 false C (21)/(1)  0 true The values in regions A and C satisfy the inequality, but we cannot include the endpoint in region C, even though the symbol is , since it will give us a 0 denominator. So the solution is the set (-, -10]  (10, ).

Example: Solve the rational inequality First we need to know when the rational expression is undefined (denominator = 0). p + 4 = 0  p = -4 Then we need to solve the equality

Example continued: Multiplying both sides of the equation by p + 4, the equality will hold when p = 3p(p + 4)  p = 3p2 + 12p  0 = 3p2 + 11p We can factor this quadratic to solve the equation. p(3p + 11) = 0  p = 0 or p = -11/3 We can then examine the four regions of the number line that are created by these three values.

Example continued: -1 -2 -3 -4 1 A B C D Region Test Point Value Substitute into factored Inequality Truth of Inequality A (-5)/(-1)  -15 false B / (-23/6)/(1/6)  -23/2 true C (-2)/(2)  -6 false D (1)/(5)  3 true

Example: The values in regions B and D satisfy the inequality, but we cannot include the left endpoint in region B, even though the symbol is , since it will give us a 0 denominator. So the solution is the set (-4, -11/3]  [0, ).

Quadratic Functions and Their Graphs
§ 8.5 Quadratic Functions and Their Graphs

Graphing Quadratic Functions
We first examined the graph of f(x) = x2 back in chapter 3. We looked at the graphs of general quadratic functions of the form f(x) = ax2 + bx + c in chapter 5. We discovered that the graph of a quadratic function is a parabola opening upward or downward, depending on the coefficient of the x2 term, a. The highest point or lowest point on the parabola is the vertex. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola.

Example: Graph f(x) = x2 Note that a = 1 in standard form. Which way does it open? What is the vertex? What is the axis of symmetry? Graph g(x) = x2 + 3 and h(x) = x2 – 3 What is the vertex of each function? What is the axis of symmetry of each function? Continued

Example continued: x y f(x) = x2 g(x) = x2 + 3 h(x) = x2 – 3

Graphing the Parabola Defined by f(x) = x2 + k If k is positive, the graph of f(x) = x2 + k is the graph of y = x2 shifted upward k units. If k is negative, the graph of f(x) = x2 + k is the graph of y = x2 shifted downward |k| units. The vertex is (0, k), and the axis of symmetry is the y-axis.

Example: Graph f(x) = x2. Graph g(x) = (x – 3)2 and h(x) = (x + 3)2. What is the vertex of each function? What is the axis of symmetry of each function? Continued

x y Example continued: f(x) = x2 g(x) = (x – 3)2 h(x) = (x + 3)2

Graphing the Parabola Defined by f(x) = (x – h)2 If h is positive, the graph of f(x) = (x – h)2 is the graph of y = x2 shifted to the right h units. If h is negative, the graph of f(x) = (x – h)2 is the graph of y = x2 shifted to the left |h| units. The vertex is (h, 0), and the axis of symmetry is the vertical line x = h.

Example: Graph f(x) = x2 Graph g(x) = (x – 2)2 + 4 What is the vertex? What is the axis of symmetry? Continued

Example continued: x y f(x) = x2 g(x) = (x – 2)2 + 4

Graphing the Parabola Defined by f(x) = (x – h)2 + k The parabola has the same shape as y = x2. The vertex is (h, k), and the axis of symmetry is the vertical line x = h..

Comparing Quadratic Functions
Example: Graph f(x) = x2 Graph g(x) = 3x2 and h(x) = (1/3)x2 How do the shapes of the graphs compare? Continued

x y Example continued: f(x) = x2 g(x) = 3x2 h(x) = (1/3)x2

Graphing the Parabola Defined by f(x) = ax2 If a is positive, the parabola open upward, and if a is negative, the parabola opens downward. If |a| > 1, the graph of the parabola is narrower than the graph of f(x) = x2. If |a| < 1, the graph of the parabola is wider than the graph of f(x) = x2.

Example: Without graphing, how does the graph of g(x) = – 4(x + 9)2 – 1 compare with the graph of f(x) = x2? First rewrite the function in the form of g(x) = –4(x – (–9))2 – 1. This new format tells us that if we start with the standard parabola, we shift 9 units to the left and one unit down. Then we flip it so that it opens down. The shape then becomes much wider. The vertex will be at (–9, –1) and the axis of symmetry is the vertical line x = –9.

Further Graphing of Quadratic Functions
§ 8.6 Further Graphing of Quadratic Functions

Graphs of Quadratic Functions
In the previous section, we discovered that if the equation of a quadratic is written in the right format, we know a lot of information about the graph If the quadratic is written in the form f(x) = a(x – h)2 + k, then we can find the vertex, axis of symmetry, whether it opens up or down, and the width.

Since we can find out a lot about a quadratic function before we ever graph it, it would be in our best interest to get the quadratic into the appropriate form so we can easily find that information. The techniques we use would be similar to completing the square.

Example: Graph f(x) = -3x2 + 6x Find the vertex, axis, and any intercepts. Before graphing, we rewrite the equation into the form that will communicate information about the graph. Continued

Example continued: The vertex is at (1,7). The axis of symmetry is the line x = 1 Continued

Example continued: Intercepts for f(x) = -3(x – 1)2 + 7 occur when x = 0 or y = 0. y-intercept occurs when we set x = 0. y = -3(0 – 1)2 + 7 y = -3(- 1)2 + 7 y = -3(1) + 7 y = 4 y-intercept is the point (0, 4). Continued

Example continued: x-intercept(s) occur when we set y = 0 0 = –3(x – 1)2 + 7 3(x – 1)2 = 7 (x – 1)2 = 7/3 Note that the square root of 7/3 is about 1.7, so x-intercepts are at about (-0.7, 0) and (2.7, 0) when we sketch the graph of the quadratic equation. x-intercepts are at the two points, and Continued

Example continued: x y The vertex is at (1,7). The axis of symmetry is the line x = 1. y-intercept is the point (0,4). x-intercepts are at the two points and

Finding a Maximum or Minimum
In many applications, you are not interested in the entire graph of the quadratic function, but merely the vertex (the highest or lowest point of the graph). In that case, it is not necessary to convert the equation of the quadratic into the previous form that gives you much more information than just the vertex. There is a formula (derived in the text, if you are interested) for finding the vertex of the parabola using standard form.

The vertex of f(x) = ax2 + bx + c, is at The maximum or minimum value of the parabola occurs at this vertex. The maximum or minimum value will be the second coordinate of the vertex.

Example: The Utah Ski Club sells calendars to raise money. The profit P, in cents, from selling x calendars is given by the equation P(x) = 360x – x2. What is the maximum profit the club will earn? Since the maximum value will occur at the vertex, we find the coordinates according to the previous formula. a = -1 and b = 360, so Continued

Example continued: The maximum profit will occur when the club sells 180 calendars. We substitute that number into the profit formula to find P(180) = 360(180) – (180)2 = – 32400 = cents (remember to read the problem carefully) = $324

Quadratic Equations and Functions

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