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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