SOLVING QUADRATIC INEQUALITEIS WORKSHEET WITH ANSWERS

Solve for x :

Problem 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

Problem 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.

Problem 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.

Problem 4 :

x² ≤ x + 2

Solution :

x² ≤ x + 2

Subtract x and 2 on both sides, we get

x² – x – 2 ≤ x + 2 – x – 2

x² – x – 2 ≤ 0

f(x) ≤ 0

By factorization, we get

(x - 2) (x + 1)  =  0

x  =  -1 and x  =  2 (critical numbers)

From the table, the possible values of x are

-1 ≤ x ≤ 2

By writing it as interval notation, we get

[-1, 2]

So, the required solution is -1 ≤ x ≤ 2

Problem 5 :

3x²- 6x + 3 > 0

Solution :

Let f(x)  =  3x²- 6x + 3

f(x) > 0

3x²- 6x + 3 > 0

By factorization, we get

x²- 2x + 1 = 0

(x-1)²  =  0

Since we have square, for all positive and negative values of x we will get a positive value.

So, the answer is all real numbers.

Problem 6 :

2x² – 5 ≤ 3x

Solution :

2x² – 5 ≤ 3x

Subtract 3x on both sides, we get

2x² – 5 – 3x ≤ 3x – 3x

2x² – 5 – 3x ≤ 0

2x² – 3x – 5 ≤ 0

By factorization, we get

(2x – 5) (x + 1)  =  0

x  =  5/2 and x  =  -1

From the table, the possible values of x are

-1 ≤ x ≤ 5/2

By writing it as interval notation, we get

 [-1, 5/2]

So, the required solution is -1 ≤ x ≤ 5/2

Problem 7 :

3x² ≥ 2(x + 4)

Solution :

3x² ≥ 2(x + 4)

3x² ≥ 2x + 8

Subtract 2x and 8 on both sides, we get

3x² – 2x - 8 ≥ 2x + 8 – 2x – 8

3x² – 2x - 8 ≥ 0

 By factorization, we get

 (3x + 4) (x – 2)  =  0

x  =  2 and x  =  -4/3

From the table, the possible values of x are

x ≤ -4/3 or x ≥ 2

By writing it as interval notation, we get

(-∞, -4/3] u [2, ∞)

So, the required solution is x ≤ -4/3 or x ≥ 2

Problem 8 :

 6(x² + 2) < 17x

Solution :

6(x² + 2) < 17x

6x² + 12 < 17x

Subtract 17x on both sides, we get

6x² + 12 – 17x < 17x – 17x

6x² + 12 – 17x < 0

6x² – 17x + 12 < 0

By factorization, we get

(2x – 3) (3x – 4)  =  0

x  =  3/2 and x  =  4/3

From the table, the possible values of x are

4/3 < x < 3/2

By writing it as interval notation, we get

(4/3, 3/2)

So, the required solution is 4/3 < x < 3/2

Problem 9 :

 9x² ≤ 12x - 4

Solution :

9x² ≤ 12x - 4

Subtract 12x and add 4 on both sides, we get

9x² – 12x + 4 ≤ 12x – 4 – 12x + 4

9x² – 12x + 4 ≤ 0

By factorization, we get

(3x – 2) (3x – 2)  =  0

x  =  2/3 

So, the possible value of x is 2/3

Problem 10 :

 4 < 5x² + 8x

Solution :

4 < 5x² + 8x

Subtract 4 on both sides, we get

4 - 4 < 5x² + 8x – 4

0 < 5x² + 8x – 4

Subtract 5x² and 8x on both sides, we get

-5x² – 8x + 0 < 5x² + 8x – 4 - 5x² – 8x

-5x² – 8x + 0 < -4

Add 4 on both sides, we get

-5x² – 8x + 0 + 4 < -4 + 4

-5x² – 8x + 4 < 0

Multiplying by ‘-‘ sign on both sides, so change the inequality < into >

5x² + 8x – 4 > 0

By factorization, we get

(5x – 2) (x + 2)  =  0

x  =  2/5 and x  =  -2

From the table, the possible values of x are

x < -2 or x > 2/5

By writing it as interval notation, we get

(-∞, -2) u (2/5, ∞)

So, the required solution is x < -2 or x > 2/5

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