SOLVING QUADRATIC EQUATIONS BY TAKING SQUARE ROOTS

Solve the equation.

Problem 1 :

x² = 289

Solution :

x² = 289

x = √289

x = √(17 ⋅ 17)

x = ±17

Problem 2 :

x² - 169 = 0

Solution :

x² - 169 = 0

x² = 169

x = √169

x = √(13 ⋅ 13)

x = ±13

Problem 3 :

2x² - 512 = 0

Solution :

2x² - 512 = 0

2x² = 512

x² = 512/2

x² = 256

x = √256

x = √(16 ⋅ 16)

x = ±16

Problem 4 :

3x² - 150 = 282

Solution :

3x² - 150 = 282

3x² = 282 + 150

3x² = 432

x² = 432/3

x² = 144

x = √(12 ⋅ 12)

x = ±12

Problem 5 :

1/2x² - 8 = 16

Solution :

1/2x² - 8 = 16

(1/2)x² = 16 + 8

(1/2)x² = 24

x² = 24 × 2

x² = 48

x = √48

x = √(4 ⋅ 4 ⋅ 3)

x = ±4√3

Problem 6 :

(2/3)x² - 4 = 12

Solution :

(2/3)x² - 4 = 12

(2/3)x² = 12 + 4

(2/3)x² = 16

x² = 24

x = √(2 ⋅ 2 ⋅ 2 ⋅ 3)

x = ±2√6

Problem 7 :

2x² + 5 = 5x² - 37

Solution :

2x² + 5 = 5x² - 37

2x² - 5x² = -37 - 5

-3x² = -42

x² = -42/-3

x² = 14

x = ±√14

Problem 8 :

4(x² - 8) = 84

Solution :

4(x² - 8) = 84

4x² - 32 = 84

4x² = 84 + 32

4x² = 116

x² = 116/4

x² = 29

x = ±√29

Problem 9 :

3(x² + 2) = 18

Solution :

3(x² + 2) = 18

3x² + 6 = 18

3x² = 18 + 6

3x² = 24

x² = 24/3

x² = 8

x = ±√(2 ⋅ 2 ⋅ 2)

x = ±2√2

Problem 10 :

2(x + 2)² = 72

Solution :

2(x + 2)² = 72

Dividing by 2 on both sides.

(x + 2)² = 36

(x + 2) = √(6 ⋅ 6)

x + 2 = ±6

x + 2 = 6 and x + 2 = -6

x = 4 and x = -8

Problem 11 :

3(x - 3)² + 2 = 26

Solution :

3(x - 3)² + 2 = 26

Subtracting 2 on both sides.

3(x - 3)² = 24

Divide by 3 on both sides.

(x - 3)² = 8

(x - 3) = √(2 ⋅ 2 ⋅ 2)

x - 3 = ±2√2

x - 3 = 2√2 and x - 3 = -2√2

x = 3 + 2√2 and x = 3 - 2√2

Problem 12 :

(3x + 2)² - 49 = 0

Solution :

(x - 3)² - 49 = 0

Adding 49 on both sides.

(x - 3)² = 49

(x - 3) = √(7 ⋅ 7)

x - 3 = ±7

x - 3 = 7 and x - 3 = -7

x = 7 + 3 and x = -7 + 3

x = 10 and x = -4