# Inequalities Worksheet

### Solutions

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, 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, +∞)