( Quadratic Inequalities

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

Quadratic Inequality


2. Represent these values in the real line. Take one point of each interval and evaluate the sign in each:

Interval

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.

Interval

S = (-∞, 2) Unión (4, ∞)


x² + 2x +1 ≥ 0

x² + 2x +1 = 0

Quadratic Inequality

(x + 1)² ≥ 0

As a number squared is always positive the solution is Any Real Number.

    Solution
x² + 2x +1 ≥ 0 (x + 1)² ≥ 0 R
x² + 2x +1 > 0 (x + 1)² > 0 R-1
x² + 2x +1 ≤ 0 (x + 1)² ≤ 0 x = − 1
x² + 2x +1 < 0 (x + 1)² < 0 vacio

x² + x +1 > 0

x² + x +1 = 0

Quadratic Inequality Solution


When there are no real roots, give the polynomial any value if:

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

The obtained sign does not coincide with that of the inequality and thus, has no solution.

  Solution
x² + x +1 ≥ 0 R
x² + x +1 > 0 R
x² + x +1 ≤ 0 vacio
x² + x +1 < 0 vacio

Examples

1 7x² + 21x − 28 < 0

x² +3x − 4 < 0

x² +3x − 4 = 0

Quadratic Inequality Operations

P(−6) = (−6)² +3 · (−6)− 4 > 0

P(0) = 0² +3 · 0 − 4 < 0

P(3) = 3² +3 · 3 − 4 > 0

Quadratic Inequality Interval

(−4, 1)

2 −x² + 4x − 7 < 0

x² − 4x + 7 = 0

Quadratic Inequality Operations

P(0) = −0² + 4 ·0 − 7 < 0

S = R

3inecuación

Quadratic Inequality Operations

Interval

P(−3) = 4 · (−3)² − 16 > 0

P(0) = 4 · 0 ² − 16 < 0

P(3) = 4 · 3 ² − 16 > 0

Interval

(-∞, −2] Unión [2, +∞)