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 |  |
Example 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)
Example 2
−x² + 4x − 7 < 0
x² − 4x + 7 = 0
P(0) = −0² + 4 ·0 − 7 < 0
S = 
Example 3
P(−3) = 4 · (−3)² − 16 > 0
P(0) = 4 · 0 ² − 16 < 0
P(3) = 4 · 3 ² − 16 > 0
(-∞, −2] ∪ [2, +∞)
Check for the best Maths tutors on Superprof.
Check for outstanding maths tutors near me here.
Get information about maths tuition in the UK.
The platform that connects tutors and students
Did you like this article? Rate it!
Loading...
