**Solving quadratic inequalities examples :**

Here we are going to see some example problems based on the concept solving quadratic inequalities.

**Solving quadratic inequalities :**

Let f(x) = ax²+ bx + c, be a quadratic function or expression. a, b, c ∈ R, a ≠ 0. Then f(x) ≥ 0, f(x) > 0, f(x) ≤ 0 and f(x) < 0 are known as quadratic inequality.

**Working Rules for solving quadratic inequality:**

**Step 1 :**

If the coefficient of x² is not positive multiply the inequality by − 1.Note that the sign of the inequality is reversed when it is multiplied by a negative quantity.

**Step 2 :**

Factorise the quadratic expression and obtain its solution by equating the linear factors to zero.

**Step 3 :**

Plot the roots on real line. The roots will divide the real line in three parts.

**Step 4 :**

In the right most part, the quadratic expression will have positive sign and in the left most part, the expression will have positive sign and in the middle part, the expression will have negative sign.

**Step 5 :**

Obtain the solution set of the given inequality by selecting the appropriate part in 4.

**Example 1 :**

Solve the inequality x² − 7x + 6 > 0

**Solution :**

x² − 7x + 6 > 0

(x − 1) (x − 6) > 0

On equating the factors to zero, we see that x = 1, x = 6 are the roots of the quadratic equation.

If we plot these points on the number line, we will get intervals (-∞, 1) (1, 6) (6, ∞).

From (-∞, 1) let us take -1 |
(-1 − 1) (-1 − 6) > 0 -2(-7) > 0 14 > 0 |

From (1, 6) let us take 4 |
(4 − 1) (4 − 6) > 0 3(-2) > 0 -6 > 0 |

From (6, ∞) let us take 7 |
(7 − 1) (7 − 6) > 0 6(1) > 0 6 > 0 |

Hence the solution set is x ∈ (− ∞, 1) ∪ (6, ∞)

**Example 2 :**

Solve the inequality -x² + 3x - 2 > 0

**Solution :**

-x² + 3x - 2 > 0

Multiplying by negative sign on both sides

x²- 3x + 2 < 0

(x − 1) (x − 2) < 0

x - 1 = 0 x - 2 = 0

x = 1 and x = 2

If we plot these points on the number line, we will get intervals (-∞, 1) (1, 2) (2, ∞).

From (-∞, 1) let us take -1 |
(x − 1) (x − 2) < 0 (-1 − 1) (-1 − 2) < 0 (-2)(-3) > 0 6 < 0 |

From (1, 2) let us take 1.5 |
(x − 1) (x − 2) < 0 (1.5 − 1) (1.5 − 2) < 0 (-0.5)(-0.5) > 0 0.25 < 0 |

From (2, ∞) let us take 7 |
(7 − 1) (7 − 6) > 0 6(1) > 0 6 > 0 False |

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