Question
Solve the following inequalities using the graphing method.
a) 2𝑥^2 − 5𝑥 − 12 ≥ 0
b) 𝑥^2 − 10𝑥 + 25 ≤ 0
a) 2𝑥^2 − 5𝑥 − 12 ≥ 0
b) 𝑥^2 − 10𝑥 + 25 ≤ 0
Answers
Answered by
GPT 3.5
To solve inequalities using the graphing method, you need to graph the inequality and find the intervals where the graph is above or below the x-axis.
a) 2𝑥^2 − 5𝑥 − 12 ≥ 0
First, find the x-intercepts of the quadratic equation by setting it equal to zero and solving for x.
2𝑥^2 − 5𝑥 − 12 = 0
(2𝑥 + 3)(𝑥 − 4) = 0
The x-intercepts are 𝑥 = -3/2 and 𝑥 = 4.
These x-intercepts divide the x-axis into three intervals: 𝑥 < -3/2, -3/2 < 𝑥 < 4, and 𝑥 > 4.
Next, plot these points on a number line and choose a test point in each interval to determine if the quadratic is positive or negative in that interval.
Test 𝑥 = -4: We substitute this value into the inequality to determine if it is true or false.
2(-4)^2 − 5(-4) − 12 ≥ 0
32 + 20 - 12 ≥ 0
40 ≥ 0
Since 40 is greater than or equal to 0, the quadratic is positive in this interval (𝑥 < -3/2).
Test 𝑥 = 0: We substitute this value into the inequality to determine if it is true or false.
2(0)^2 − 5(0) − 12 ≥ 0
0 - 0 - 12 ≥ 0
-12 < 0
Since -12 is less than 0, the quadratic is negative in this interval (-3/2 < 𝑥 < 4).
Test 𝑥 = 5: We substitute this value into the inequality to determine if it is true or false.
2(5)^2 − 5(5) − 12 ≥ 0
50 - 25 - 12 ≥ 0
13 ≥ 0
Since 13 is greater than or equal to 0, the quadratic is positive in this interval (𝑥 > 4).
Therefore, the solution to the inequality 2𝑥^2 − 5𝑥 − 12 ≥ 0 is 𝑥 < -3/2 or 𝑥 > 4.
b) 𝑥^2 − 10𝑥 + 25 ≤ 0
To solve this inequality, find the x-intercepts of the quadratic equation.
𝑥^2 − 10𝑥 + 25 = 0
(𝑥 − 5)^2 = 0
𝑥 = 5
The x-intercept is 𝑥 = 5.
Since the quadratic is a perfect square, it touches the x-axis at 𝑥 = 5, but does not cross it. Therefore, the graph of the quadratic lies entirely below the x-axis.
The solution to the inequality 𝑥^2 − 10𝑥 + 25 ≤ 0 is 𝑥 = 5.
a) 2𝑥^2 − 5𝑥 − 12 ≥ 0
First, find the x-intercepts of the quadratic equation by setting it equal to zero and solving for x.
2𝑥^2 − 5𝑥 − 12 = 0
(2𝑥 + 3)(𝑥 − 4) = 0
The x-intercepts are 𝑥 = -3/2 and 𝑥 = 4.
These x-intercepts divide the x-axis into three intervals: 𝑥 < -3/2, -3/2 < 𝑥 < 4, and 𝑥 > 4.
Next, plot these points on a number line and choose a test point in each interval to determine if the quadratic is positive or negative in that interval.
Test 𝑥 = -4: We substitute this value into the inequality to determine if it is true or false.
2(-4)^2 − 5(-4) − 12 ≥ 0
32 + 20 - 12 ≥ 0
40 ≥ 0
Since 40 is greater than or equal to 0, the quadratic is positive in this interval (𝑥 < -3/2).
Test 𝑥 = 0: We substitute this value into the inequality to determine if it is true or false.
2(0)^2 − 5(0) − 12 ≥ 0
0 - 0 - 12 ≥ 0
-12 < 0
Since -12 is less than 0, the quadratic is negative in this interval (-3/2 < 𝑥 < 4).
Test 𝑥 = 5: We substitute this value into the inequality to determine if it is true or false.
2(5)^2 − 5(5) − 12 ≥ 0
50 - 25 - 12 ≥ 0
13 ≥ 0
Since 13 is greater than or equal to 0, the quadratic is positive in this interval (𝑥 > 4).
Therefore, the solution to the inequality 2𝑥^2 − 5𝑥 − 12 ≥ 0 is 𝑥 < -3/2 or 𝑥 > 4.
b) 𝑥^2 − 10𝑥 + 25 ≤ 0
To solve this inequality, find the x-intercepts of the quadratic equation.
𝑥^2 − 10𝑥 + 25 = 0
(𝑥 − 5)^2 = 0
𝑥 = 5
The x-intercept is 𝑥 = 5.
Since the quadratic is a perfect square, it touches the x-axis at 𝑥 = 5, but does not cross it. Therefore, the graph of the quadratic lies entirely below the x-axis.
The solution to the inequality 𝑥^2 − 10𝑥 + 25 ≤ 0 is 𝑥 = 5.
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