Quadratic Inequalities
Consider the inequality:
x2 − 6x + 8 > 0
It can be solved by following these steps:
1.Equal the polynomial of the first member to zero and the roots of the quadratic equation are obtained.
x2 − 6x + 8 = 0
2. Represent these values in the real line. Take one point of each interval and evaluate the sign in each:
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
3. The solution is defined by the intervals (or the interval) that have the same sign as the polynomial.
S = (-∞, 2) 
(4, ∞)
x2 + 2x +1 ≥ 0
x2 + 2x +1 = 0
(x + 1)2 ≥ 0
As a number squared is always positive the solution is 
.
| Solution | ||
|---|---|---|
| x2 + 2x +1 ≥ 0 | (x + 1)2 ≥ 0 | |
| x2 + 2x +1 > 0 | (x + 1)2 > 0 |  |
| x2 + 2x +1 ≤ 0 | (x + 1)2 ≤ 0 | x = − 1 |
| x2 + 2x +1 < 0 | (x + 1)2 < 0 |  |
x2 + x +1 > 0
x2 + x +1 = 0
When there are no real roots, give the polynomial any value if:
The sign obtained coincides with the inequality, the solution is .
The obtained sign does not coincide with that of the inequality and thus, has no solution.
| Solution | |
|---|---|
| x2 + x +1 ≥ 0 | |
| x2 + x +1 > 0 | |
| x2 + x +1 ≤ 0 | |
| x2 + x +1 < 0 | |
Examples
1 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)
2 −x2 + 4x − 7 < 0
x2 − 4x + 7 = 0
P(0) = −02 + 4 ·0 − 7 < 0
S =
3
P(−3) = 4 · (−3)2 − 16 > 0
P(0) = 4 · 0 2 − 16 < 0
P(3) = 4 · 3 2 − 16 > 0
(-∞, −2] [2, +∞)
