Quadratic Inequalities
Consider the inequality:
x^{2} − 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.
x^{2} − 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) = 0^{2 } − 6 · 0 + 8 > 0
P(3) = 3^{2 } − 6 · 3 + 8 = 17 − 18 < 0
P(5) = 5^{2 } − 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, ∞)
x^{2} + 2x +1 ≥ 0
x^{2} + 2x +1 = 0
(x + 1)^{2} ≥ 0
As a number squared is always positive the solution is .
Solution | ||
---|---|---|
x^{2} + 2x +1 ≥ 0 | (x + 1)^{2} ≥ 0 | |
x^{2} + 2x +1 > 0 | (x + 1)^{2} > 0 | |
x^{2} + 2x +1 ≤ 0 | (x + 1)^{2} ≤ 0 | x = − 1 |
x^{2} + 2x +1 < 0 | (x + 1)^{2} < 0 |
x^{2} + x +1 > 0
x^{2} + 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 | |
---|---|
x^{2} + x +1 ≥ 0 | |
x^{2} + x +1 > 0 | |
x^{2} + x +1 ≤ 0 | |
x^{2} + x +1 < 0 |
Examples
1 7x^{2} + 21x − 28 < 0
x^{2} +3x − 4 < 0
x^{2} +3x − 4 = 0
P(−6) = (−6)^{2} +3 · (−6)− 4 > 0
P(0) = 0^{2} +3 · 0 − 4 < 0
P(3) = 3^{2} +3 · 3 − 4 > 0
(−4, 1)
2 −x^{2} + 4x − 7 < 0
x^{2} − 4x + 7 = 0
P(0) = −0^{2} + 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, +∞)