EXAMPLES OF SOLVING QUADRATIC INEQUALITIES

Solve for x :

Example 1 :

3x² ≤ 12

Solution :

3x² ≤ 12

Subtract 12 on both sides, we get

3x² - 12 ≤ 12 – 12

3x² - 12 ≤ 0

Let f(x) ≤ 0

3x² – 12  =  0

3x²  =  12

x²  =  4

So, x  =  2 and x  =  -2 (critical numbers)

The critical numbers are dividing the number line into three intervals.

From the table, the possible values of x are

-2 ≤ x ≤ 2

By writing it as interval notation, we get

[-2, 2]

So, the required solution is -2 ≤ x ≤ 2

Example 2 :

x² – 4x + 4 < 0

Solution :

Let f(x)  =  x² – 4x + 4

f(x) < 0

x² – 4x + 4 < 0

By factorization, we get

x² – 2x(2) + 2² < 0

(x-2)² < 0

For any values of x, f(x) will not give negative value. So the solution is all real values.

Example 3 :

2x² + 7x < -6

Solution :

2x² + 7x < -6

Add 6 on both sides, we get

2x² + 7x + 6 < -6 + 6

2x² + 7x + 6 < 0

Let f(x)  =  2x² + 7x + 6

f(x) < 0

2x² + 7x + 6 < 0

By factorization, we get

(2x + 3) (x + 2)  =  0

x  =  -3/2 and x  =  -2 (critical numbers)

From the table, the possible values of x are

-2 < x < -3/2

By writing it as interval notation, we get

(-2, -3/2)

So, the required solution is -2 < x < -3/2

Example 4 :

x² + 6x + 8 < 0

Solution :

x² + 6x + 8 < 0

x² + 2x + 4x + 8 < 0

x(x + 2) + 4(x + 2) < 0

(x + 4)(x + 2) < 0

Getting critical numbers,

x + 4 = 0 and x + 2 = 0

x = -4 and x = -2

Decomposing into intervals, we get

(-∞, -4) (-4, -2) and (-2, ∞)

f(x) = (x + 4)(x + 2) < 0

Applying one of the value from the interval (-∞, -4)

Let x = -5

f(-5) = (-5 + 4)(-5 + 2)

= -1(-3)

= 3 < 0

It is false. (-∞, -4) is not a solution.

Applying one of the value from the interval (-4, -2)

Let x = -3

f(-3) = (-3 + 4)(-3 + 2)

= 1(-1)

= -1 < 0

It is true. (-4, -2) is a solution.

Applying one of the value from the interval (-2, ∞)

Let x = 0

f(0) = (0 + 4)(0 + 2)

= 4(2)

= 8 < 0

It is false. Then, (-2, ∞) is not a solution.

So, the required solution is -4 ≤ x ≤ -2.

Example 5 :

When a baseball is hit by a batter, the height of the ball, ℎ(𝑡), at time 𝑡, is determined by the equation

ℎ(𝑡) = −16𝑡² + 64𝑡 + 4

For which interval of time is the height of the ball greater than or equal to 52 feet?

Solution :

Given condition is when the height will be ≥ 52

−16𝑡² + 64𝑡 + 4 ≥ 52

−16𝑡² + 64𝑡 + 4 - 52 ≥ 0

−16𝑡² + 64𝑡  - 48 ≥ 0

Dividing by -16, we get

𝑡² - 4𝑡  + 3 ≤ 0

(t - 1)(t - 3) ≤ 0

Finding the critical numbers, we get

t = 1 and t = 3

Decomposing into intervals, we get

(-∞, 1) (1, 3) and (3, ∞)

Applying one of the value from the interval (-∞, 1)

Let x = 0

f(0) = (0 - 1)(0 - 3)

= 3

It is false, then (-∞, 1) is not a solution.

Applying one of the value from the interval (1, 3)

Let x = 2

f(2) = (2 - 1)(2 - 3)

= -1

It is true, then (1, 3) is a solution.

Applying one of the value from the interval (3, ∞)

Let x = 4

f(4) = (4 - 1)(4 - 3)

= 3(1)

= 3

Then the required solution is 1 < x < 3.

Example 6 :

The surface area, A, of a cylinder with radius r is given by the formula 𝐴 = 2𝑟² − 5𝑟. What possible radii would result in an area that is greater than 12 cm² ?

Solution :

A = 2𝑟² − 5𝑟

The required area should be greater than 12, then 

2𝑟² − 5𝑟 > 12

2𝑟² − 5𝑟 - 12 > 0

2𝑟² − 8𝑟 + 3r - 12 > 0

2r(r - 4) + 3(r - 4) > 0

(2r + 3)(r - 4) > 0

Finding critical number, we get

r = -3/2 and r = 4

Here negative value is not possible for radius, then the required radius should be greater than 4.

Example 7 :

When a projectile is fired into the air, its height h, in metres, t seconds later is given by the equation

ℎ(𝑡) = 11𝑡 − 3𝑡²

When is the projectile at least 6 m above the ground?

Solution : 

The given condition is that the height is atleast 6 m, then 

11𝑡 − 3𝑡² > 6

− 3𝑡² + 11t - 6 > 0

Multiply by negative, we get 

3𝑡² - 11t + 6 < 0

3𝑡² - 2t - 9t + 6 < 0

t(3t - 2) - 3(3t - 2) < 0

(t - 3)(3t - 2) < 0

t = 3 and t = 2/3

So, the required solution is 2/3 < t < 3.

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