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

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

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

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

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].

Example 5 :

Solve the inequality 3x² − 5x − 1 < 4x² + 7x + 19

Solution :

3x² − 5x − 1 < 4x² + 7x + 19

3x² - 4x² − 5x - 7x - 1 - 19 < 0

-x² − 12x - 20 < 0

Multiplying by negative, we get

x² + 12x + 20 > 0

(x + 10)(x + 2) > 0

The critical points are -10 and -2.

The intervals are (-∞, -10) (-10, -2) and (-2, ∞)

x ∈ (-2, ∞)

Say x = 0

(0 + 10)(0 + 2) > 0

10(2) > 0 (True)

So, the solutions are (-∞, -10) and  (-2, ∞).

Example 6 :

Solve 2x² + 13x – 17 ≥ x² + 4x + 5.

Solution :

2x² + 13x – 17 ≥ x² + 4x + 5

2x² - x² + 13x - 4x - 17 - 5 < 0

x² − 9x - 22 < 0

Multiplying by negative, we get

x² − 9x - 22 < 0

x² − 11x + 2x - 22 < 0

x(x - 11) + 2(x - 11) < 0

(x - 11)(x + 2) < 0

The critical points are -2 and 11

The intervals are (-∞, -2) (-2, 11) and (11, ∞)

x ∈ (11, ∞)

Say x = 12

(12 - 11)(12 + 2) < 0

1(14) < 0

14 < 0 (False)

So, the solution is (-2, 11).