Inequalities Worksheet

1 Solve:

2Solve:

4x² − 4x + 1 ≤ 0

3Solve:

4Calculate the values of k for which the roots of the equation x² − 6x + k = 0 are two real and distinct numbers.

5Solve:

1

2

3

6Solve:

7Solve:

1

2x⁴ − 25x² − 144 < 0

3x⁴ − 16x² − 225 ≥ 0  

1

Solve:

2

Solve:

4x² − 4x + 1 ≤ 0

4x² − 4x + 1 = 0

3

Solve:

     

    The numerator is always positive.

   

The denominator cannot be zero.

Therefore, the original inequality will be equivalent to:

x² − 4 > 0

(−-∞ , −2) (2, +∞)

4

Calculate the values of k for which the roots of the equation x² − 6x + k = 0 are two real and distinct numbers.

(−6)² − 4k > 0

36 − 4k > 0          − 4k > − 36        k < 9

(−∞, 9)

5

Solve:

1

x = 4

y = 2

2

x + y = 0        (0, 0)     (1, -1)

2 + 2 ≥ 0

2x − y = 0      (0, 0)     (1, 2)

2 ·2 − 2 ≥ 0

3

x + y = 0        (0, 0)     (1, -1)

2 + 2 ≥ 0

2x − y = 0      (0, 0)     (1, 2)

2 ·2 − 2 ≥ 0

2 ≤ 6

     

6

Solve:

(x +1) · 10 + x ≤ 6 (2x + 1)

10x + 10 + x ≤ 12 x + 6

10 x + x - 12x ≤ 6 - 10

−x ≤ − 4       x ≥ 4

[4, 7)

7

Solve:

1

As the first factor is always positive, consider the sign of the 2nd factor.

P(−17) = (−17) ² + 12 · 17 − 64 > 0

P(0) = 0² + 12 · 0 − 64 < 0

P(5) = 5 ² + 12 · 5 − 64 > 0

(-∞, −16] [4, ∞)

2x⁴ − 25x² − 144 < 0

x⁴ − 25x² − 144 = 0

(−4, −3) (−3, 3 ) (3, 4) .

3x⁴ − 16x² − 225 ≥ 0 

x⁴ − 16x² − 225 = 0 

(x² - 25) · (x² + 9) ≥ 0

The second factor is always positive and nonzero, therefor, only consider the sign of the 1st factor.

(x² − 25) ≥ 0

(-∞, −5] [5, +∞)