# EXAMPLES OF SOLVING QUADRATIC INEQUALITIES

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].

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