**Graph and solve quadratic inequalities worksheet :**

Here we are going to see some practice questions on graph and solving quadratic inequalities.

(1) Solve for "x" and graph the solution

x² - 7x + 6 > 0

(2) Solve for "x" and graph the solution

-x² + 3x - 2 > 0

(3) Solve for "x" and graph the solution

4x² - 25 ≥ 0

(4) Solve for "x" and graph the solution

2x² − 12x + 50 ≤ 0

**Question 1 :**

Solve for "x" and graph the solution

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. Plotting these roots on real line and marking positive and negative alternatively from the right most part we obtain the corresponding number line as

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, ∞)

**Question 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

On equating the factors to zero, we see that x = 1, x = 2 are the roots of the quadratic equation. Plotting these roots on real line and marking positive and negative alternatively from the right most part we obtain the corresponding number line as

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 |

**Hence the solution set is (1, 2)**

**Question 3 :**

Solve the inequality 4x² - 25 ≥ 0

**Solution :**

4x² - 25 ≥ 0

(2x)² - 5² ≥ 0

(2x - 5) (2x + 5) ≥ 0

2x - 5 ≥ 0 (or) 2x + 5 ≥ 0

x = 5/2 (or) x = -5/2

On equating the factors to zero, we see that x = 5/2, x = -5/2 are the roots of the quadratic equation. Plotting these roots on real line and marking positive and negative alternatively from the right most part we obtain the corresponding number line as

If we plot these points on the number line, we will get intervals (-∞, -5/2) (-5/2, 5/2) (5/2, ∞).Graph solutions to quadratic inequalities

From (-∞, -5/2) let us take -3 |
(2x - 5) (2x + 5) ≥ 0 (-6 − 5) (-6 + 5) < 0 (-11)(-1) > 0 11 > 0 |

From (-5/2, 5/2) let us take 0 |
(2x - 5) (2x + 5) ≥ 0 (0− 5) (0 + 5) < 0 (-5)(5) > 0 -25 > 0 |

From (5/2, ∞) let us take 4 |
(2x - 5) (2x + 5) ≥ 0 (8 − 5) (8 + 5) < 0 (3)(13) > 0 39 > 0 |

**Hence the solution set is **(-∞, -5/2) U (5/2, ∞)

**Question 4 :**

Solve for "x" and graph the solution

2x² − 12x + 50 ≤ 0

**Solution :**

2x² − 12x + 50 ≤ 0

2(x²− 6x + 25) ≤ 0

x² − 6x + 25 ≤ 0

(x²− 6x + 9) + 25 − 9 ≤ 0

(x − 3)² + 16 ≤ 0

This is not true for any real value of x.

Given inequation has no solution.

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