Solving Quadratic Equations by Factoring Worksheet

Problem 1 :

x2 + 11x + 28 = 0

a. x = -7and -4                    c. x = -4 and 7

b. x = -7 and 4                    d. x = 4 and 7

Solution

Problem 2 :

6x2 - 4x + 8

a. 6x(3x – 2)          c. 6x2 – 4x + 8

b. 3(2x2 – 4x + 8)         d. 2(3x2 – 2x + 4)

Solution

Problem 3 :

-12x2 - 8x = 0

a. x = 0 and x = 12          c. x = 0 and x = -2

b. x = 0 and x = -2/3          d. x = 0 and x = -3/2

Solution

Problem 4 :

Solve by factoring.

4x2 + 10x – 24 = 0

a.  3/2, -1         c. 4, -1

b. -4, 3/2          d. -4, 4

Solution

Problem 5 :

6x2 = 42

a. -√7, √42            c. √7, -√7

b. -√42/6, √42/6           d. √7

Solution

1)  x = -7 and x = -4

2)  2(3x2 – 2x + 4)

3)  x = 0 and x = -2/3

4)  x = 3/2 , x = -4

5)  x = ±√7

Solve the equation using square root property.

Problem 1 :

x² = 289

Solution

Problem 2 :

x² - 169 = 0

Solution

Problem 3 :

2x² - 512 = 0

Solution

Problem 4 :

3x² - 150 = 282

Solution

Problem 5 :

1/2x² - 8 = 16

Solution

Problem 6 :

(2/3)x² - 4 = 12

Solution

Problem 7 :

2x² + 5 = 5x² - 37

Solution

Problem 8 :

4(x² - 8) = 84

Solution

Problem 9 :

3(x² + 2) = 18

Solution

Problem 10 :

2(x + 2)² = 72

Solution

Problem 11 :

3(x - 3)² + 2 = 26.

Solution

Problem 12 :

(3x + 2)² - 49 = 0

Solution

 1)  x = ±172)  x = ±133)  x = ±164)  x = ±125)  x = ±4√36)  x = ±2√6 7)  x = ±√148)  x = ±√299)  x = ±2√210)  x = 4 and x = -811)  x = 3 + 2√2 and x = 3 - 2√212)  x = 10 and x = -4

Problem 1 :

Two numbers are such that thrice the smaller number exceeds twice the greater one by 18 and 1/3 of the smaller and 1/5 of the greater number are together 21. The numbers are -

(a) (45,36)     (b) (50, 38)      (c) (54,45)       (d) (55,41)

Solution

Problem 2 :

Two sides of an equilateral triangle are shortened by 12 units 13 units and 14 units respectively and a right angle is formed. The sides of the equilateral triangle is

(a) 17 units      (b) 16 units      (c) 15 units      (d) 18 units

Solution

Problem 3 :

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4cm. The sides are

(a) (11 cm, 15 cm)       (b) (12 cm, 16 cm)

(c) (20 cm, 24 cm)       (d) None of these

Solution

Problem 4 :

The sum of two numbers is 45 and the mean proportional between them is 18. The numbers are

(a) (15, 30)     (b) (32, 13)     (c) (36, 9)        (d) (25, 20)

Solution

1) (45,36)

2)  17 units.

3) 12 and 16 are the required sides.

4)  x = 9 and x = 36

5)  x = 64/113 and x = 1/2

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