Solve for the following quadratic inequalities.
Example 1 :
y² - y ≥ 12
Solution :
y² - y ≥ 12
Subtract 12 on both sides, we get
y² - y - 12 ≥ 12 - 12
y² - y - 12 ≥ 0
y² - y - 12 = 0
(y + 3) (y - 4) = 0
y + 3 = 0 and y - 4 = 0
y = -3 and y = 4
Writing them as intervals, we get
(-∞, -3] [-3, 4] [4, ∞)
Applying any values within the interval, we get
Intervals |
signs of factors (y + 3) (y - 4) |
given
inequality |
(-∞, -3] Say y = -4 |
(-) (-) |
(+) |
[-3, 4] Say y = 0 |
(+) (-) |
(-) |
[4, ∞) Say y = 5 |
(+) (+) |
(+) |
From the above table, we come to know that the interval [-3, 4] satisfies the given inequality.
Hence, the solution is [-3, 4].
Example 2 :
-y² + 4 > 0
Solution :
-y² + 4 > 0
Let f(y) > 0
-y² = -4
y = 2
So, y = 2 and y = -2
Writing them as intervals, we get
(-∞, -2) (-2, 2) (2, ∞)
Applying any values within the interval, we get
Intervals |
signs of factors f(y) |
given inequality |
(-∞, -2) Say y = -3 |
(-) (-) |
(+) |
(-2, 2) Say y = 0 |
(+) (-) |
(-) |
(2, ∞) Say y = 3 |
(+) (+) |
(+) |
From the above table, we come to know that the interval (-2, 2) satisfies the given inequality.
Hence, the solution is (-2, 2).
Example 3 :
-y² + 3y - 2 ≤ 0
Solution :
First let us solve the given quadratic equation by factoring.
The coefficient of y must be positive, so we have to multiply the inequality by negative.
y² - 3y + 2 ≤ 0
Multiply the equation by negative.
y² - 3y + 2 = 0
(y - 1) (y - 2) = 0
y - 1 = 0 y - 2 = 0
y = 1 y = 2
Writing them as intervals, we get
(-∞, 1] [1, 2] [2, ∞)
Applying any values within the interval, we get
Intervals |
signs of factors (y - 1) (y - 2) |
given inequality |
(-∞, -1] Say y = 0 |
(-) (-) |
(+) |
[1, 2] Say y = 1.5 |
(-) (+) |
(-) |
[2, ∞) Say y = 3 |
(+) (+) |
(+) |
From the above table, we come to know that the interval [1, 2] satisfies the given inequality.
Hence, the solution is [1, 2].
Example 4 :
-2y² + 5y + 12 ≤ 0
Solution :
First let us solve the given quadratic equation by factoring.
The coefficient of y must be positive, so we have to multiply the inequality by negative.
2y² - 5y - 12 = 0
2y² - 8y + 3y - 12 = 0
2y(y - 4) + 3(y - 4) = 0
(2y + 3) (y - 4) = 0
2y + 3 = 0 y - 4 = 0
y = -3/2 y = 4
Writing them as intervals, we get
(-∞, -3/2] [-3/2, 4] [4, ∞)
Applying any values within the interval, we get
Intervals |
signs of factors (2y + 3) (y - 4) |
given inequality |
(-∞, -3/2] Say y = -2 |
(-) (-) |
(+) |
[-3/2, 4] Say y = 0 |
(+) (-) |
(-) |
[4, ∞) Say y = 5 |
(+) (+) |
(+) |
From the above table, we come to know that the interval [-3/2, 4] satisfies the given inequality.
Hence, the solution is [-3/2, 4].
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM