SOLVING QUADRATIC INEQUALITIES

Solve for the following quadratic inequalities :

Problem 1 :

5 - 4y - y² > 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 + 4y - 5 < 0

y² + 4y - 5 = 0

(y - 1) (y + 5) = 0

  y - 1 = 0 and y + 5 = 0

y = 1 and y = -5

Writing them as intervals, we get

(-∞, -5) (-5, 1) (1, ∞)

Applying any values within the interval, we get

From the above table, we come to know that the interval (-5, 1) satisfies the given inequality.

Hence, the solution is (-5, 1).

Problem 2 :

1 - y - 2y² < 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 + y - 1 < 0

Multiply the equation by negative.

2y² + y - 1 = 0

2y² + 2y - y - 1 = 0

2y(y + 1) - 1(y + 1) = 0

(2y - 1) (y + 1) = 0

2y - 1 = 0 and y + 1 = 0

2y = 1 and y = -1

   y = 1/2 and y = -1

Writing them as intervals, we get

(-∞, -1) (-1, 1/2) (1/2, ∞)

Applying any values within the interval, we get

From the above table, we come to know that the interval (-1, 1/2) satisfies the given inequality.

Hence, the solution is (-1, 1/2).

Problem 3 :

(y - 3) (y + 2) > 0

Solution :

(y - 3) (y + 2) = 0

  y - 3 = 0 and y + 2 = 0

y = 3 and y = -2

Writing them as intervals, we get

(-∞, -2) (-2, 3) (3, ∞)

Applying any values within the interval, we get

From the above table, we come to know that the interval (-2, 3) satisfies the given inequality.

Hence, the solution is (-2, 3).

Problem 4 :

y² - 2y - 3 < 0

Solution :

y² - 2y - 3 = 0

y² + y - 3y - 3 = 0

y(y + 1) - 3(y + 1) = 0

(y + 1) (y - 3) = 0

y + 1 = 0 and y - 3 = 0

y = -1 and y = 3

Writing them as intervals, we get

(-∞, -1) (-1, 3) (3, ∞)

Applying any values within the interval, we get

From the above table, we come to know that the interval (-1, 3) satisfies the given inequality.

Hence, the solution is (-1, 3).

Problem 5 :

Find the set of values of x for which x² − 2x − 24 < 0 and 12 − 5x ≥ x + 9

Solution :

Solving x² − 2x − 24 < 0 :

x² − 2x − 24 = 0

x² − 6x + 4x − 24 = 0

(x - 6) (x + 4) = 0

x = 6 an x = -4

The intervals are (-∞, -4) (-4, 6) and (6, ∞)

x ∈ (6, ∞), say x = 7

(7 - 6)(7 + 4) < 0

1(11) < 0

11 < 0 (False)

So, the solution for the given quadratic inequality is (-4, 6).

Solving 12 − 5x ≥ x + 9 :

-5x - x ≥ 9 - 12

-6x ≥ -3

x ≤ 1/2

The solution is (-∞, 1/2).

The intersection for these two intervals is (-∞, 1/2). The intersection interval of these two solutions is (-4, 1/2].

Problem 6 :

Find the set of values of x for which x² − 100 > 0 and x² + 8x − 105 > 0

Solution :

Solving x² − 100 > 0 :

x² − 100 = 0

(x - 10)(x + 10) = 0

x = -10 and x = 10

The critical points are -10 and 10

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

x ∈ (10, ∞), say x = 11

(x - 10)(x + 10) > 0

(11 - 10)(11 + 10) > 0

1(21) > 0

21 > 0 (True)

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

Solving x² + 8x − 105 > 0 :

x²  + 8x - 105 = 0

x²  + 15x - 7x - 105 = 0

x(x + 15) - 7(x + 15) = 0

(x - 7)(x + 15) = 0

x = 7 and x = -15

The critical points are -15 and 7

The intervals are (-∞, -15) (-15, 7) and (7, ∞).

x ∈ (7, ∞), say x = 8

(x - 7)(x + 15) > 0

(8 - 7)(8 + 15) > 0

1(23) > 0

23 > 0 (True)

So, the solutions are (-∞, -15) and (7, ∞).

Comparing the intervals (-∞, -10) (10, ∞) and  (-∞, -15) (7, ∞) the intersection parts is (-∞, -15) and (10, ∞).