Quadratic Inequalities

Quadratic Inequalities
Sovling Quadratic Inequalities Quadratic Inequalities Solving Quadratic Inequalities Quadratic Inequalities This module explores methods for solving quadratic inequalities. The stated goal is the finding of the solution set for the inequality. We emphasize simple graphical representation in order to help understand both the question being posed in the problem and the method of solution. Before taking on quadratic inequalities, we briefly review graphical solution of linear inequalities for comparison. The method is virtually the same for both. Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities
Inequalities Review Sovling Quadratic Inequalities General Inequalities Consider: f(x) ≤ c or f(x) ≥ c Questions to ask: 1. Is the inequality true for ALL x ? 2. Is the inequality true for NO x ? 3. Is the inequality true for SOME x but not others ? … or f(x) < c … or f(x) > c Quadratic Inequalities: Standard Form We keep in mind what we are looking for, which has a direct bearing on the types of solutions we can expect. The solution sets range from all real numbers to no real numbers with the possibility of some (but not all) real numbers. While we may in general think of the constant c as any real number, it is often convenient to arrange the inequality so that c = 0. 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Inequalities Review Sovling Quadratic Inequalities General Inequalities Questions with answers: 1. Is the inequality true for ALL x ? Solution set: R 2. Is the inequality true for NO x ? Solution set: { } 3. Is the inequality true for SOME x but not others ? Solution set: { x  f(x) ≤ c } Quadratic Inequalities: Standard Form We keep in mind what we are looking for, which has a direct bearing on the types of solutions we can expect. The solution sets range from all real numbers to no real numbers with the possibility of some (but not all) real numbers. While we may in general think of the constant c as any real number, it is often convenient to arrange the inequality so that c = 0. 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Examples Sovling Quadratic Inequalities Example 1 Solve: Find x such that True for ALL x Solution Set : y x y(x) = –2 ≤ 0 –2 ≤ 0 y(x) = –2 Linear Inequalities: Examples 1 and 2 In these examples we compare simple linear functions against the constant 0. In the first example we identify one side of the inequality as a function y(x) and the other side as the constant (or constant function) y = 0. We ask for which x is it true that the graph of the constant function y(x) = -2 is below the graph of y = 0, i.e. below the x-axis. Clearly this is true for all x, making the solution set the set of all real numbers R, or in interval notation (-, ). In the second example we ask for which x is it true that the constant function y(x) = -2 is greater than or equal than the constant function y = 0. In other words, for which x is the graph of y(x) = -2 above the graph of y = 0 ? Clearly this is not true for any x, making the solution set empty. Hence the solution set is { } or ∅, both of which are symbols for the empty set. R or (– , ) 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Examples Sovling Quadratic Inequalities Example 2 Solve: Find x such that True for NO x Solution Set : { } y x y(x) = –2 ≥ 0 –2 ≥ 0 y(x) = –2 Linear Inequalities: Examples 1 and 2 In these examples we compare simple linear functions against the constant 0. In the first example we identify one side of the inequality as a function y(x) and the other side as the constant (or constant function) y = 0. We ask for which x is it true that the graph of the constant function y(x) = -2 is below the graph of y = 0, i.e. below the x-axis. Clearly this is true for all x, making the solution set the set of all real numbers R, or in interval notation (-, ). In the second example we ask for which x is it true that the constant function y(x) = -2 is greater than or equal than the constant function y = 0. In other words, for which x is the graph of y(x) = -2 above the graph of y = 0 ? Clearly this is not true for any x, making the solution set empty. Hence the solution set is { } or ∅, both of which are symbols for the empty set. or O 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Examples Sovling Quadratic Inequalities Example 3 Solve: First find x such that y x y = 3 – 2x ≥ 0 3 – 2x = 0 x 3 2 = Then No match! x > 3 2 x 3 2 > For  Quadratic Inequalities: Example 3 We demonstrate the solution of the inequality graphically and symbolically by first finding the points of equality. Clearly, as a linear equation, there is one and only one solution, namely x = 3/2. This amounts to using the horizontal intercept (or x-intercept) method of solution described in Section 2.4. Then we consider all x > 3/2 and work backward, to see what inequality this produces, with 0 on the right. This gives us 3 – 2x < 0, which is NOT the given inequality. So, evidently all x > 3/2 fail to satisfy the original inequality. Now we consider all x < 3/2 and work backward to the inequality with 0 on the right. This gives us 3 – 2x < 0, which is exactly the original inequality. Thus the solution set is { x | x ≤ 3/2 } = (-, 3/2]. Note that the interval is closed on the right, i.e. 3/2 is included, since the original inequality is ≤ and not <. Graphically we think of a function y = 3 – 2x and find the horizontal intercept at x = 3/2. For the inequality to be satisfied, the graph of y must be above the line y = 0 (i.e., the x=axis). Projecting the points on this portion of the graph onto the x-axis gives us the solution set – almost. We still have to account for the point of intersection – do we include it or not? Since the original inequality is a non-strict inequality, we include the point of intersection, which means we include 3/2 in the solution set. 3 2 we have 2x > 3 Subtracting 2x , y = 3 – 2x 0 > 3 – 2x … or 3 – 2x < 0 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Examples Sovling Quadratic Inequalities Example 3 Solve: (continued) y x y = 3 – 2x ≥ 0 x 3 2 > For No match x 3 2 < But for Match! we have 2x < 3 x < 3 2 x > 3 2 Subtracting 2x ,  Quadratic Inequalities: Example 3 We demonstrate the solution of the inequality graphically and symbolically by first finding the points of equality. Clearly, as a linear equation, there is one and only one solution, namely x = 3/2. This amounts to using the horizontal intercept (or x-intercept) method of solution described in Section 2.4. Then we consider all x > 3/2 and work backward, to see what inequality this produces, with 0 on the right. This gives us 3 – 2x < 0, which is NOT the given inequality. So, evidently all x > 3/2 fail to satisfy the original inequality. Now we consider all x < 3/2 and work backward to the inequality with 0 on the right. This gives us 3 – 2x < 0, which is exactly the original inequality. Thus the solution set is { x | x ≤ 3/2 } = (-, 3/2]. Note that the interval is closed on the right, i.e. 3/2 is included, since the original inequality is ≤ and not <. Graphically we think of a function y = 3 – 2x and find the horizontal intercept at x = 3/2. For the inequality to be satisfied, the graph of y must be above the line y = 0 (i.e., the x=axis). Projecting the points on this portion of the graph onto the x-axis gives us the solution set – almost. We still have to account for the point of intersection – do we include it or not? Since the original inequality is a non-strict inequality, we include the point of intersection, which means we include 3/2 in the solution set. 3 2 0 < 3 – 2x … or 3 – 2x > 0 y = 3 – 2x Solution Set: = 3 2 x | x { } ≤ ( , ] – 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Absolute Values in Inequalities
Sovling Quadratic Inequalities Recall: Basic Absolute Value Facts  x  ≥ 0 for all real x  x  =  –x  for all real x Absolute Values in Inequalities We recall the definition of absolute value to remind ourselves that when we talk about any real number a, we leave open the possibility that a might be either positive or negative. The first two facts are then quite obvious: |x| is never negative, regardless of the value of x itself, x and –x have exactly the same absolute value. The next two facts are a bit less obvious, which is why the animations show what happens if x is positive as well as what happens if x is negative. Assuming x is not zero, the trivial case, we start by placing |x| on the positive side of 0 and less than the given number b. We see that b must be positive so that –b is surely negative is shown in its position relative to the other numbers. If we remove the absolute value, then x might be positive and –x negative, as shown in the animation. Or, as the animation shows, x might be negative and –x positive, in which case |–x| is located exactly where |x| is located. This follows from the above properties since |x| and |–x| are equal. Clearly, then whether x is positive or negative it lies between –b and b. The last animated fact shows that if |x| is larger than a positive number b, then either x > b (if x is positive) or x < –b (if x is negative). Here again, the animation leads us through the possibility that x is positive and hence x > b, and then through the possibility that x is negative, so that –x > b, making x < –b. Again we note that |–x| and |x| are equal, placing them at the same location on the number line. 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Absolute Values in Inequalities
Linear Inequalities Sovling Quadratic Inequalities Basic Absolute Value Facts If  x  < b then –b < x < b If  x  > b > 0 WHY ? ? ? –b –x x  –x  x | |  x  –x x b Absolute Values in Inequalities We recall the definition of absolute value to remind ourselves that when we talk about any real number a, we leave open the possibility that a might be either positive or negative. The first two facts are then quite obvious: |x| is never negative, regardless of the value of x itself, x and –x have exactly the same absolute value. The next two facts are a bit less obvious, which is why the animations show what happens if x is positive as well as what happens if x is negative. Assuming x is not zero, the trivial case, we start by placing |x| on the positive side of 0 and less than the given number b. We see that b must be positive so that –b is surely negative and is shown in its position relative to the other numbers. If we remove the absolute value, then x might be positive and –x negative, as shown in the animation. Or, as the animation shows, x might be negative and –x positive, in which case |–x| is located exactly where |x| is located. This follows from the above properties since |x| and |–x| are equal. Clearly, then whether x is positive or negative it lies between –b and b. The last animated fact shows that if |x| is larger than a positive number b, then either x > b (if x is positive) or x < –b (if x is negative). Here again, the animation leads us through the possibility that x is positive and hence x > b, and then through the possibility that x is negative, so that –x > b, making x < –b. Again we note that |–x| and |x| are equal, placing them at the same location on the number line. then either x > b or x < -b WHY ? ? ? b –b –x x  –x  –x  x   x  x 7/9/2013 Quadratic Inequalities 9 Section v5.0 Quadratic Inequalities 9 7/9/2013 2/6/2013

Absolute Values & Quadratics
Sovling Quadratic Inequalities Example 4: Now where is x ? Either x ≥ 2 | x | ≥ 2 x –2 2  x  x ] [ OR x ≤ –2 Solution set: { x | x ≤ -2 } {x | x ≥ 2 } ∪ Example 4: Absolute Values in Inequalities Here we deal with absolute values in inequalities. We begin with a very simple inequality: | x | ≥ 2 and ask where x can be. As shown, x can be either greater than +2 OR x might be negative and less than -2. That is, x must lie outside the interval (-2, 2). Thus the solution set for this inequality is { x | x ≤ -2 }  { x | x ≥ 2 } OR (-, -2]  [2, ). Note the solution set includes 2 and -2 so the intervals are closed intervals. To connect absolute values to quadratic forms we ask: where would x2 be? Clearly, if x ≥ 2 then x is certainly positive, so x2 ≥ 2x ≥ 2(2) = 4, giving x2 ≥ 4. However, if x ≤ -2 then x is clearly negative, so that when both sides of this inequality are multiplied by either x or -2 we get x2 ≥ -2x and -2x ≥ (-2)(-2) = 4 We conclude that x2 ≥ 4 for x ≤ -2. This allows us to say with confidence that x2 – 4 ≥ 0 if and only if | x | ≥ 2. This gives us solution set { x |  x  ≥ 2 }. OR (- , -2] [2, ) ∪ { x |  x  ≥ 2 } OR So how does this connect to quadratics ? 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Example 4: | x | ≥ 2 (continued) x 2 –2 ] [ x2 So how does this connect to quadratics ? If x ≤ –2 < 0 then x is negative, and x2 = x(x) ≥ x(–2) > 0 WHY ? Example 4: Absolute Values in Inequalities Here we deal with absolute values in inequalities. We begin with a very simple inequality: | x | ≥ 2 and ask where x can be. As shown, x can be either greater than +2 OR x might be negative and less than -2. That is, x must lie outside the interval (-2, 2). Thus the solution set for this inequality is { x | x ≤ -2 }  { x | x ≥ 2 } OR (-, -2]  [2, ). Note the solution set includes 2 and -2 so the intervals are closed intervals. To connect absolute values to quadratic forms we ask: where would x2 be? Clearly, if x ≥ 2 then x is certainly positive, so x2 ≥ 2x ≥ 2(2) = 4, giving x2 ≥ 4. However, if x ≤ -2 then x is clearly negative, so that when both sides of this inequality are multiplied by either x or -2 we get x2 ≥ -2x and -2x ≥ (-2)(-2) = 4 We conclude that x2 ≥ 4 for x ≤ -2. This allows us to say with confidence that x2 – 4 ≥ 0 if and only if | x | ≥ 2. This gives us solution set { x |  x  ≥ 2 }. x ≤ –2 < 0 also implies (–2)x ≥ (–2)(–2) = (–2)2 = 4 Hence for x ≤ –2 , x2 ≥ (–2)x ≥ (–2)(–2) = 4 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Example 4: | x | ≥ 2 (continued) x2 x 2 –2 ] [  x  Similarly, for x ≥ 2 , we have x2 = x(x) ≥ 2x and 2x ≥ 2(2) = 4 WHY ? … and thus x2 ≥ 4 Example 4: Absolute Values in Inequalities Here we deal with absolute values in inequalities. We begin with a very simple inequality: | x | ≥ 2 and ask where x can be. As shown, x can be either greater than +2 OR x might be negative and less than -2. That is, x must lie outside the interval (-2, 2). Thus the solution set for this inequality is { x | x ≤ -2 }  { x | x ≥ 2 } OR (-, -2]  [2, ). Note the solution set includes 2 and -2 so the intervals are closed intervals. To connect absolute values to quadratic forms we ask: where would x2 be? Clearly, if x ≥ 2 then x is certainly positive, so x2 ≥ 2x ≥ 2(2) = 4, giving x2 ≥ 4. However, if x ≤ -2 then x is clearly negative, so that when both sides of this inequality are multiplied by either x or -2 we get x2 ≥ -2x and -2x ≥ (-2)(-2) = 4 We conclude that x2 ≥ 4 for x ≤ -2. This allows us to say with confidence that x2 – 4 ≥ 0 if and only if | x | ≥ 2. This gives us solution set { x |  x  ≥ 2 }. Hence x2 – 4 ≥ 0 if and only if x ≥ 2 OR x ≤ –2 … that is, for | x | ≥ 2 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities y x Example 5 Solve: First solve If | x | > 2 , No Match ! y = x2 – 4 ≤ 0 y = x2 – 4 x2 – 4 = 0 Then x2 = 4 and thus y > 0 y > 0 x = ±  4 = ± 2  ● then x2 > 22 = 4 Quadratic Inequalities: Example 5 Note here that the inequality features a 0 on one side. This is important here, since we will use the square root property to solve the associated equation. This is done to locate the boundary points for the inequality. Once the boundary points are found, we check the intervals on either side to see the effect on the functional expression whether or not these points satisfy the original inequality. In the example we see that the points where the expression x2 – 4 changes from negative to positive, or vice versa, are at x = –2 and x = 2. A simple way of saying we want to focus on values of x less than –2 and greater than 2 is to check for all x with  x  > 2. When we check these values of x we find that the expression x2 – 4 is greater than 0, which is not compatible with the original inequality. Hence, we conclude the solutions (if there are any) are not in the regions where  x  > 2. We next check for values of x between the boundary points – that is for all x with –2 < x 2 (i.e. for  x  < 2). This produces the expression x2 – 4 which is now less than 0 and this is compatible with the original inequality. The only thing left to do is to check the boundary points themselves. Since the original inequality is a non-strict inequality (less than or equal) the boundary points are included in the solution set. (–2, 0) (2, 0) … or x2 > (– 2)2 = 4 Either way y = x2 – 4 > 0 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities y x Example 5 Solve: If | x | > 2 (continued) y = x2 – 4 ≤ 0 Match ! y = x2 – 4 y = x2 – 4 > 0 If | x | < 2 then –2 < x < 2 and  ● Quadratic Inequalities: Example 5 Note here that the inequality features a 0 on one side. This is important here, since we will use the square root property to solve the associated equation. This is done to locate the boundary points for the inequality. Once the boundary points are found, we check the intervals on either side to see the effect on the functional expression whether or not these points satisfy the original inequality. In the example we see that the points where the expression x2 – 4 changes from negative to positive, or vice versa, are at x = –2 and x = 2. A simple way of saying we want to focus on values of x less than –2 and greater than 2 is to check for all x with  x  > 2. When we check these values of x we find that the expression x2 – 4 is greater than 0, which is not compatible with the original inequality. Hence, we conclude the solutions (if there are any) are not in the regions where  x  > 2. We next check for values of x between the boundary points – that is for all x with –2 < x 2 (i.e. for  x  < 2). This produces the expression x2 – 4 which is now less than 0 and this is compatible with the original inequality. The only thing left to do is to check the boundary points themselves. Since the original inequality is a non-strict inequality (less than or equal) the boundary points are included in the solution set. (–2, 0) (2, 0) x2 < 4 so that x2 – 4 < 0 y < 0 Including boundary points, Solution Set { x | –2 ≤ x ≤ 2 } = [ –2 , 2 ] 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities y x Example 5 -- Revisited Find boundary points for First solve: Splits domain into three intervals Test a point x in each interval y = x2 – 4 y = x2 – 4 ≤ 0 y > 0 y > 0 x2 – 4 = 0 So x2 = 4 x = ±  4 = ± 2 Quadratic Inequalities: Example 5 Revisited We solve the associated equation to locate the boundary points for the inequality. These points define three disjoint (non-overlapping) open intervals. We test each interval by testing one point in each interval. We need test only one point since the function y = x2 – 4 is continuous over the interval (in fact, over its entire domain) and its graph crosses the x-axis only at the zeros of the function, (the boundary points for the inequality). Hence, between the zeros of the function it cannot have value 0, that is, the graph cannot cross the x-axis between the zeros. Thus the sign of the functional values cannot change between zeros. The table thus shows that points in each of the left and right intervals do not represent solutions of the inequality, and that points in the central interval (–2 , 2) all represent solutions – since 0 is a solution and since the function is continuous, all the points in the interval represent solutions. We must then check the boundary points themselves. We see that both –2 and +2 are solutions. Putting these together with the open interval (–2 , 2) we see that the closed interval [–2 , 2] represents the solution set. (–2, 0) (2, 0) y < 0 WHY ? 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities y x Example 5 Revisited Test a point x in each interval (continued) y = x2 – 4 Interval x Inequality True/ False (– , –2) –4 (–4)2 – 4 ≤ 0 False ∞ (–2 , 2) (0)2 – 4 ≤ 0 TRUE ( 2 , ) (4)2 – 4 ≤ 0 False ∞ (–4, 0) (4, 0) ● ● ● ● ● Quadratic Inequalities: Example 5 Revisited We solve the associated equation to locate the boundary points for the inequality. These points define three disjoint (non-overlapping) open intervals. We test each interval by testing one point in each interval. We need test only one point since the function y = x2 – 4 is continuous over the interval (in fact, over its entire domain) and its graph crosses the x-axis only at the zeros of the function, (the boundary points for the inequality). Hence, between the zeros of the function it cannot have value 0, that is, the graph cannot cross the x-axis between the zeros. Thus the sign of the functional values cannot change between zeros. The table thus shows that points in each of the left and right intervals do not represent solutions of the inequality, and that points in the central interval (–2 , 2) all represent solutions – since 0 is a solution and since the function is continuous, all the points in the interval represent solutions. We must then check the boundary points themselves. We see that both –2 and +2 are solutions. Putting these together with the open interval (–2 , 2) we see that the closed interval [–2 , 2] represents the solution set. (–2, 0) (2, 0) Test boundary points x = –2 : (–2)2 – 4 ≤ 0 x = 2 : (2)2 – 4 ≤ 0 Solution set: { x │–2 ≤ x ≤ 2 } = [ –2, 2 ] 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Example 6 Solve: Rewrite: Find x such that y x x2 + 4 ≥ 2x + 3 y = x2 – 2x + 1 x2 – 2x + 1 ≥ 0 y = x2 – 2x + 1 = (x – 1)2 = 0 Quadratic Inequalities: Example 6 Here we have a more complex quadratic. The obvious first step, as in solving quadratic equations, is to move all terms to one side of the inequality. By convention (i.e. the way most people are used to seeing the expression) we move all terms to the left side of the inequality, leaving 0 on the right side. We could certainly move everything to the right side and solve just as easily. We first find the points of equality; that is, where is the expression on the left equal to that on that on the right? In this case we factor the quadratic expression into two identical factors, that is, into a perfect square. Using either the zero-product property or the square root property, we easily see that the expression is zero if and only if x = 1. Since the expression is a square it is either positive or zero, meaning it is greater than or equal to zero for all values of x. The solution set is thus the set of all real numbers R, or in interval notation it is the open interval (-, ). If we change the ≥ to = , then we find the solution set is { 1 }. If we change the ≥ to < , then we find the solution set is { } = ∅. and thus x = 1 For x ≠ 1 y = (x – 1)2 > 0 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Example 6 Solve: (continued) y x x2 + 4 ≥ 2x + 3 y = x2 – 2x + 1 For x ≠ 1 y = (x – 1)2 > 0 Including the boundary point y > 0 y > 0 Solution set: ● Quadratic Inequalities: Example 6 Here we have a more complex quadratic. The obvious first step, as in solving quadratic equations, is to move all terms to one side of the inequality. By convention (i.e. the way most people are used to seeing the expression) we move all terms to the left side of the inequality, leaving 0 on the right side. We could certainly move everything to the right side and solve just as easily. We first find the points of equality; that is, where is the expression on the left equal to that on that on the right? In this case we factor the quadratic expression into two identical factors, that is, into a perfect square. Using either the zero-product property or the square root property, we easily see that the expression is zero if and only if x = 1. Since the expression is a square it is either positive or zero, meaning it is greater than or equal to zero for all values of x. The solution set is thus the set of all real numbers R, or in interval notation it is the open interval (-, ). If we change the ≥ to = , then we find the solution set is { 1 }. If we change the ≥ to < , then we find the solution set is { } = ∅. = ( , ) – R 1 Question: What if x2 + 4 < 2x + 3 ? What if x2 + 4 = 2x + 3 ? 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities y x Example 7 Solve: y = –(x – 4)2 – 1 ≥ 0 Note that for any x,  –(x – 4)2 ≤ 0 (4, –1) Thus –(x – 4)2 – 1 ≤ –1 < 0 So, NO (real) x satisfies Quadratic Inequalities: Example 7 We start by recalling that, unless otherwise specified, we shall deal only with real numbers without recourse to the complex number system. Here we see a polynomial with all negative terms and ask for which values of x it is greater than or equal to 0. Clearly, with all negative terms, the expression cannot be greater than 0. So we ask whether it can be equal to 0. We note that even when (x – 4)2 is zero, the constant term -1 makes the expression negative. We conclude that the given quadratic expression is always less than 0 and hence is not greater than or equal to 0 for any real value of x. Hence the solution set is empty, that is { } or ∅. We also note that y = -(x – 4)2 – 1 is in vertex form and has the vertex of the graph at (4, -1). Since the lead coefficient is -1, the parabola opens downward and is thus the entire graph is below the vertex and hence below the x-axis. There are no x-intercepts for the graph, i.e. no points that would correspond to solutions of the quadratic equation. If we consider the very different inequality -(x – 4 )2 – 1 ≤ 0 we see that this one is satisfied for all x, that is, the solution set is R. y = –(x – 4)2 – 1 ≥ 0 Solution set: { } OR O Question: What if y = –(x – 4)2 – 1 ≤ 0 ? 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Example 8 Solve: Factoring the difference of squares Factors must have the same sign x2 – 5 ≥ 0 (x – )(x ) ≥ 0  WHY ? So, Quadratic Inequalities: Example 8 In this example we have a quadratic expression in the form of a difference of squares (since 5 is the square of ). There are two approaches here. The first approach is to factor into conjugate binomials, as shown. Then apply the zero-product property. Since this is not an equation, we have the possibility that the product is greater than 0 – which simply tells us that the two factors have the same sign. We then proceed to evaluate each factor as positive, followed by the same computations for each factor being negative. Each analysis produces a compound inequality with overlapping solution sets. The final result is that either or giving the solution set The second approach is to rewrite the inequality as x2 ≥ 5 and then apply the absolute value property (see Example 4) to show that x2 – 5 ≥ 0 if and only if  x  ≥ A slightly different approach here is to note that using the principal root since the absolute value cannot be negative. Then means that either x ≥ 5 or x ≥ –5 as we have already seen in earlier examples. For equations the square-root property tells us that if x2 = 5 then x = ± However, if x2 > 5 it does not follow that x > ± since x might be negative and smaller than For example, if x = –3 then x2 – 5 = 9 – 5 = 4 > 0, but it is not true that –3 > Thus we cannot use the square-root property. (x – ) ≥ 0  and (x ) ≥ 0  OR (x – ) ≤ 0  and (x ) ≤ 0  7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Exercise Solve: Factor out 3 , rewrite as an equivalent inequality Factoring again: There are several ways to solve 3x2 + 3x – 18 > 0 x2 + x – 6 > 0 (x – 3)(x + 2) > 0 Exercise Method 1 uses boundary points and test points. Change the inequality to an equation with 0 on the right side. Solve for values of x that make this true. These are the boundary points where the graph crosses the x-axis. Check the sign of the function at points to each side of each boundary point to which intervals produce positive values (the ones we want in this example) and negative values (where our inequality is NOT true). The positive intervals (in this example) form the solution set. Boundary points are excluded from the solution set if and only if the inequality is a strict inequality (which is the case in this example). Method 2 takes note of the fact that the product of the two factors is positive, implying that the signs of the factors are the same – either both positive or both negative. Assume each case (both positive, then both negative) and see what each pair of inequalities produces. Each gives half the solution set. Method 3 indicates graphing the expression as a function of x and noting which parts of the graph are above the x-axis. The projection of these parts of the graph onto the x-axis form the solution set. Method 4 lends itself to a graphing calculator. Enter the expression into the function screen and use the Trace key to graph the function. Access the table generated and look for values of 0 for the function. These are the boundary points and can be used in conjunction with the graph to determine the solution set, i.e. the regions where the expression is positive. 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Exercise Solve: There are several ways to solve 1. Find boundary points and test adjacent intervals 2. Note signs of factors, solve in pairs 3. Solve graphically 4. Solve by building tables 3x2 + 3x – 18 > 0 Exercise Method 1 uses boundary points and test points. Change the inequality to an equation with 0 on the right side. Solve for values of x that make this true. These are the boundary points where the graph crosses the x-axis. Check the sign of the function at points to each side of each boundary point to which intervals produce positive values (the ones we want in this example) and negative values (where our inequality is NOT true). The positive intervals (in this example) form the solution set. Boundary points are excluded from the solution set if and only if the inequality is a strict inequality (which is the case in this example). Method 2 takes note of the fact that the product of the two factors is positive, implying that the signs of the factors are the same – either both positive or both negative. Assume each case (both positive, then both negative) and see what each pair of inequalities produces. Each gives half the solution set. Method 3 indicates graphing the expression as a function of x and noting which parts of the graph are above the x-axis. The projection of these parts of the graph onto the x-axis form the solution set. Method 4 lends itself to a graphing calculator. Enter the expression into the function screen and use the Trace key to graph the function. Access the table generated and look for values of 0 for the function. These are the boundary points and can be used in conjunction with the graph to determine the solution set, i.e. the regions where the expression is positive. Solution Set: { x  x < –2 }  { x  x > 3 } = (– , –2)  (3, ) 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Exercise Solve: Add 1 and rewrite as an equivalent inequality There are several ways to solve 1. Find boundary points and test adjacent intervals 2. Note signs of factors and solve in pairs x2 – 10 < –1 x2 – 9 = x2 – 32 < 0 (x – 3)(x + 3) < 0 Exercise Method 1 uses boundary points and test points. Change the inequality to an equation with 0 on the right side. Solve for values of x that make this true. These are the boundary points where the graph crosses the x-axis. Check the sign of the function at points to each side of each boundary point to which intervals produce positive values (the ones we want in this example) and negative values (where our inequality is NOT true). The positive intervals (in this example) form the solution set. Boundary points are excluded from the solution set if and only if the inequality is a strict inequality (which is the case in this example). Method 2 notes that the product of the two factors is negative, implying that the signs of the factors are different – one positive, one negative. Assume each case and see what each pair of inequalities produces. Each gives half the solution set. Care should be used in thinking about the square root property, since x2 < 9 implies both x < 3 and x > –3 (NOT x < –3 ). However, if we note that we see that –3 < x < 3, as already shown. Method 3 indicates graphing the expression as a function of x and noting which parts of the graph are above the x-axis. The projection of these parts of the graph onto the x-axis form the solution set. Method 4 lends itself to a graphing calculator. Enter the expression into the function screen and use the Trace key to graph the function. Access the table generated and look for values of 0 for the function. These are the boundary points and can be used in conjunction with the graph to determine the solution set, i.e. the regions where the expression is negative. 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Sovling Quadratic Inequalities Exercise Solve: 3. Solve graphically 4. Solve by building tables x2 – 10 < –1 (x – 3)(x + 3) < 0 Question: Exercise Method 1 uses boundary points and test points. Change the inequality to an equation with 0 on the right side. Solve for values of x that make this true. These are the boundary points where the graph crosses the x-axis. Check the sign of the function at points to each side of each boundary point to which intervals produce positive values (the ones we want in this example) and negative values (where our inequality is NOT true). The positive intervals (in this example) form the solution set. Boundary points are excluded from the solution set if and only if the inequality is a strict inequality (which is the case in this example). Method 2 notes that the product of the two factors is negative, implying that the signs of the factors are different – one positive, one negative. Assume each case and see what each pair of inequalities produces. Each gives half the solution set. Care should be used in thinking about the square root property, since x2 < 9 implies both x < 3 and x > –3 (NOT x < –3 ). However, if we note that we see that –3 < x < 3, as already shown. Method 3 indicates graphing the expression as a function of x and noting which parts of the graph are above the x-axis. The projection of these parts of the graph onto the x-axis form the solution set. Method 4 lends itself to a graphing calculator. Enter the expression into the function screen and use the Trace key to graph the function. Access the table generated and look for values of 0 for the function. These are the boundary points and can be used in conjunction with the graph to determine the solution set, i.e. the regions where the expression is negative. Can we use the square root property ? Note: For x2 < 9 we have x2  = | x | < 9  = 3 and we know | x | < 3 means –3 < x < 3 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Think about it ! 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Examples Sovling Quadratic Inequalities Example 3 Solve: Find x such that Thus For so 0 > 3 – 2x so 0 < 3 – 2x Solution set: y x Match ! y = 3 – 2x ≥ 0 y = 3 – 2x 3 – 2x = 0 x 3 2 = No match? x 3 2 > we have 2x > 3 Include 3 2 ... or 3 – 2x < 0 x < 3 2 x > 3 2 x 3 2 < we have 2x < 3 Quadratic Inequalities: Example 3 We demonstrate the solution of the inequality graphically and symbolically by first finding the points of equality. Clearly, as a linear equation, there is one and only one solution, namely x = 3/2. This amounts to using the horizontal intercept (or x-intercept) method of solution described in Section 2.4. Then we consider all x > 3/2 and work backward, to see what inequality this produces, with 0 on the right. This gives us 3 – 2x < 0, which is NOT the given inequality. So, evidently all x > 3/2 fail to satisfy the original inequality. Now we consider all x < 3/2 and work backward to the inequality with 0 on the right. This gives us 3 – 2x < 0, which is exactly the original inequality. Thus the solution set is { x | x ≤ 3/2 } = (-, 3/2]. Note that the interval is closed on the right, i.e. 3/2 is included, since the original inequality is ≤ and not <. Graphically we think of a function y = 3 – 2x and find the horizontal intercept at x = 3/2. For the inequality to be satisfied, the graph of y must be above the line y = 0 (i.e., the x=axis). Projecting the points on this portion of the graph onto the x-axis gives us the solution set – almost. We still have to account for the point of intersection – do we include it or not? Since the original inequality is a non-strict inequality, we include the point of intersection, which means we include 3/2 in the solution set.  3 2 ... or 3 – 2x > 0 { x  } x 3 2 ≤ ( , ] 3 2 = – 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

Absolute Values and Quadratics
Sovling Quadratic Inequalities Example 4: Now where is x ? Either x ≥ 2 OR x ≤ -2 How does this connect to quadratics ? x ≤ -2 < 0 implies x is negative, so x2 ≥ (-2)x x ≤ -2 < 0 also implies (-2)x ≥ (-2)(-2) = (-2)2 Hence for x ≤ -2 , x2 ≥ (-2)x ≥ (-2)2 = 4 Similarly for x ≥ 2 we have x2 ≥ 2x and 2x ≥ 2(2) = 4 ... and thus x2 ≥ 4 Then x2 – 4 ≥ 0 if and only if x ≥ 2 OR x ≤ -2 ... that is, for  x  ≥ 2 x –2 2  x   x  x x2 ] [ Solution set: { x | x ≤ -2 }  {x | x ≥ 2 } OR (- , -2]  [2, ) OR { x |  x  ≥ 2 } > 0 = 4 , i.e. x2 ≥ 4 Example 4: Absolute Values in Inequalities Here we deal with absolute values in inequalities. We begin with a very simple inequality: | x | ≥ 2 and ask where x can be. As shown, x can be either greater than +2 OR x might be negative and less than -2. That is, x must lie outside the interval (-2, 2). Thus the solution set for this inequality is { x | x ≤ -2 }  { x | x ≥ 2 } OR (-, -2]  [2, ). Note the solution set includes 2 and -2 so the intervals are closed intervals. To connect absolute values to quadratic forms we ask: where would x2 be? Clearly, if x ≥ 2 then x is certainly positive, so x2 ≥ 2x ≥ 2(2) = 4, giving x2 ≥ 4. However, if x ≤ -2 then x is clearly negative, so that when both sides of this inequality are multiplied by either x or -2 we get x2 ≥ -2x and -2x ≥ (-2)(-2) = 4 We conclude that x2 ≥ 4 for x ≤ -2. This allows us to say with confidence that x2 – 4 ≥ 0 if and only if | x | ≥ 2. This gives us solution set { x |  x  ≥ 2 }.  x  ≥ 2 7/9/2013 Quadratic Inequalities Quadratic Inequalities 7/9/2013

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