### Solutions

1x2 − 6x + 8 > 0

2x2 + 2x +1 ≥ 0

3x2 + x +1 > 0

47x2 + 21x − 28 < 0

5 −x2 + 4x − 7 < 0

6

74x2 − 4x + 1 ≤ 0

8

9x4 − 25x2 + 144 < 0

10x4 − 16x2 − 225 ≥ 0

## 1

x2 − 6x + 8 > 0

x2 − 6x + 8 = 0

P(0) = 02 − 6 · 0 + 8 > 0

P(3) = 32 − 6 · 3 + 8 = 17 − 18 < 0

P(5) = 52 − 6 · 5 + 8 = 33 − 30 > 0

S = (-∞, 2) (4, ∞)

## 2

x2 + 2x +1 = 0

(x + 1)2 ≥ 0

As a number squared is always positive.

S =

## 3

x2 + x +1 > 0

x2 + x +1 = 0

P(0) = 0 + 0 + 1 > 0

The sign obtained coincides with the inequality, the solution is .

## 4

7x2 + 21x − 28 < 0

x2 +3x − 4 < 0

x2 +3x − 4 = 0

P(−6) = (−6)2 +3 · (−6)− 4 > 0

P(0) = 02 +3 · 0 − 4 < 0

P(3) = 32 +3 · 3 − 4 > 0

(−4, 1)

## 5

−x2 + 4x − 7 < 0

x2 − 4x + 7 = 0

P(0) = −02 + 4 ·0 − 7 < 0

S =

## 6

P(−3) = 4 · (−3)2 − 16 > 0

P(0) = 4 · 0 2 − 16 < 0

P(3) = 4 · 3 2 − 16 > 0

(-∞ , −2] [2, +∞)

4x2 − 4x + 1 ≤ 0

4x2 − 4x + 1 = 0

## 8

The first factor is always positive.

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, ∞)

## 9

x4 − 25x2 + 144 < 0

x4 − 25x2 + 144 = 0

(−4, −3) (3, 4) .

## 10

x4 − 16x2 − 225 ≥ 0

x4 − 16x2 − 225 = 0

(x2 - 25) · (x2 + 9) ≥ 0

The second factor is always positive and different to 0.

(x2 − 25) ≥ 0

(-∞, −5] [5, +∞)