GRAPH AND SOLVE QUADRATIC INEQUALITIES WORKSHEET

Problem 1 :

Graph and solve for x :

x² - 7x + 6 > 0

Problem 2 : 

Graph and solve for x : 

 -x² + 3x - 2 > 0

Problem 3 :

Graph and solve for x :

4x² - 25   ≥   0

Problem 4 :

Graph and solve for x :

2x² − 12x + 50 ≤ 0

Solutions

Problem 1 :

Graph and solve for x :

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, ∞).

Hence, the solution set is 

(− ∞, 1) ∪ (6, ∞)

Problem 2 : 

Graph and solve for x : 

 -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, ∞).

Hence, the solution set is 

(1, 2)

Problem 3 :

Graph and solve for x :

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

Hence, the solution set is 

(-∞, -5/2) U (5/2, ∞)

Problem 4 :

Graph and solve for x :

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.

So, the given quadratic inequality has no solution.

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