Question

Solve each equation. Some of them may have radicals in their solutions.

Answer 1

Answer: a) 3 and -3 b) 4 and 2

c) 3.5 and 2.5 d) 5.5 and 0.5 e) This is due to the difference in the coefficient of (x-3)² and the value at the right side of each equation.

Step-by-step explanation:

a) 3x²-9 = 0

3x² = 0+9

3x² = 9

Dividing both sides by 3

x² = 9/3

x² = 3

x = √3

x = +3 and -3

b) (x-3)² = 1

Taking square root of both sides to remove the square,

√(x-3)² = √1

x-3 = √1

x-3 = 1 and x-3 = -1 (note that √1 is +1 and -1)

x = 4 and x = 2

c) 4(x-3)² = 1

Dividing both sides by 4 we have;

(x-3)² = 1/4

Taking the square root of both sides to eliminate the square.

√(x-3)² = √1/4

x-3 = √1/4

x-3 = +1/2 and x-3 = -1/2

x = 1/2+3 and x = -1/2+3

x = 3.5 and 2.5

d) 2(x-3)² = 12

Dividing both sides by 2 we have;

(x-3)² = 12/2

Taking the square root of both sides to eliminate the square.

√(x-3)² = √12/2

x-3 = √6

x-3 = 2.5 and x-3 = -2.5

x = 2.5+3 and x = -2.5+3

x = 5.5 and 0.5

e) The answers to questions 1 to 4 has different solutions due to the different value of the coefficient of (x-3)² and also the value at the right hand side of the equations also differs, hence, the reason for the difference in their answers.