Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.5.1, AII.5.3, AII.6.3
A projectile's height can be modeled by h = −16t2 + v0t + h0 where v0 is the projectile's initial velocity and h0 is the projectile's initial height. If it is launched straight up at 30 ft/s from 10 feet above the ground, how much time will the projectile be higher than 20 feet? This question in a quadratic inequality.
Solving quadratic inequalities is similar to solving quadratic equations, but extra steps are needed to find the range of solutions. With linear inequalities such as x > 5, there are only two possibilities: either x is bigger than 5 or smaller than 5. Obviously this one is bigger than 5. Quadratic inequalities can have 0, 1, 2, or 3 regions that might be possible solutions. Each of those regions must be tested to determine the solutions.
To solve quadratic inequalities in one variable,
Solve x2 + 7x + 12 > 0.
Solution
One side of the inequality is already zero.
Try factoring to find the zeros.
x2 + 7x + 12 > 0
(x + 3)(x + 4) > 0
x + 3 = 0 or x + 4 = 0
x = −3 or x = −4
Graph these on a number line. Use open circles because the inequality is not equal to.
Notice these points split the number line into three intervals: x < −4, −4 < x < −3, and x > −3. Pick a test point in each interval. Test −5, −3.5, 0 by substituting these points into the original inequality.
Test −5:
(−5)2 + 7(−5) + 12 > 0
2 > 0 → TRUE
Test −3.5:
(−3.5)2 + 7(−3.5) + 12 > 0
−0.25 > 0 → False
Test 0:
(0)2 + 7(0) + 12 > 0
12 > 0 → TRUE
The solution are the intervals that were true.
x < −4 or x > −3
Solve x2 − 2x ≤ 15.
Solution
Make one side of the inequality zero.
x2 − 2x ≤ 15
x2 − 2x − 15 ≤ 0
Try factoring to find the zeros.
(x − 5)(x + 3) ≤ 0
x − 5 = 0 or x + 3 = 0
x = 5 or x = −3
Graph these on a number line. Use filled circles because the inequality is equal to.
Notice these points split the number line into three intervals: x < −3, −3 < x < 5, and x > 5. Pick a test point in each interval. Test −4, 0, 6 by substituting these points into the original inequality.
Test −4:
(−4)2 − 2(−4) ≤ 15
24 ≤ 15 → False
Test 0:
(0)2 − 2(0) ≤ 15
0 ≤ 15 → TRUE
Test 6:
(6)2 − 2(6) ≤ 15
24 ≤ 15 → False
The solution are the intervals that were true.
−3 ≤ x ≤ 5
A graph can be used to help determine the regions which are solutions to the quadratic inequality. Remember the method of solving an equation by graphing? If the equation equaled zero, the solutions are the x-intercepts because those are the places where the y-value is zero. When the graph is above the x-axis, the y-values are greater than zero. When the graph is below the x-axis, the y-vales are less than zero. Thus, a quick way to determine the intervals that are solutions to the inequality it to draw a quick sketch of the graph of the parabola on the number line. The x-intercepts are the zeros, and the leading coefficient tells whether the graph opens up or down.
To use a graph to determine the solution intervals,
Solve 2x2 + 5x − 12 ≥ 0.
Solution
One side of the inequality is already zero.
Try factoring to find the zeros.
2x2 + 5x − 12 ≥ 0
(2x − 3)(x + 4) ≥ 0
2x − 3 = 0 or x + 4 = 0
2x = 3 or x = −4
x = \(\frac{3}{2}\) or x = −4
Graph these on a number line. Use filled circles because the inequality is equal to.
To determine the solution intervals, sketch a graph of the quadratic. The leading coefficient is positive, so it opens up and passes through the points on the number line.
The inequality is 2x2 + 5x − 12 ≥ 0 which is the function is greater than zero. That means the solution is the intervals where the function is above the number line (above the x-axis where the y-values are greater than zero).
The solutions are x ≤ −4 or x ≥ \(\frac{\mathbf{3}}{\mathbf{2}}\).
Solve 3x2 + 5x < 4.
Solution
Make one side of the inequality zero.
3x2 + 5x < 4
3x2 + 5x − 4 < 0
Try factoring to find the zeros.
That does not work, so use the quadratic formula to find the zeros.
$$ x=\frac{-b+\pm\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{-5\pm\sqrt{5^2-4\left(3\right)\left(-4\right)}}{2\left(3\right)} $$
$$ x=\frac{-5\pm\sqrt{73}}{6} $$
x ≈ 0.591 or −2.257
Graph these on a number line. Use open circles because the inequality is not equal to.
To determine the solution intervals, sketch a graph of the quadratic. The leading coefficient of 3x2 + 5x − 4 is positive, so it opens up.
The inequality is 3x2 + 5x − 4 < 0 which is the function is less than zero. That means the solution is the intervals where the function is below the number line (below the x-axis where the y-values are less than zero).
The solutions are −2.257 < x < 0.591
140 #27, 29, 31, 33, 35, 37, 39, 41, 43, 49, Mixed Review = 15
Mixed Review