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).