Question

Solve each equation, if possible. Write irrational numbers in simplest radical form. Describe the strategy you used to get your solution and tell why you chose that strategy. x^2+4=0 Problem 1 Hint: directions say if possible (so as you solve think about a step that you get to that prevents you from solving for the final answer). Explain your thoughts. x^2+x-6=0 Problem 2 Hint: Factor the equation and then solve to get 2 answers using the zero product property x^2-6x+7=0 Problem 3 Hint: factor using the completing the square method

Answer 1

Problem 1: x^2 + 4 = 0 , x^2 = -4 , this problem has no real solution because x = 2i which is an imaginary number

problem 2: x^2 +x -6 = 0 ,  (x + 3)(x - 2) =0 , x = 2 or x = -3

problem 3: x^2 -6x +7 =0 , (x^2 -6x +9) - 2 =0 , (x - 3)^2 = 2 , x-3 = √2 , x = 3+√2  or x - 3 = -√2 , x = 3 - √2

Answer 2

1)
x^2 + 4 = 0      Subtract 4 from each side
x^2 = -4           
You cannot square a number to get a negative number. No real solution. (If you have learned about imaginary numbers, answer is 2i)

2)
x^2+x-6=0        Factor. 3 and -2 add up to 1, and multiply to -6
(x+3)(x-2) = 0   
Zero Product Property: Which numbers would put a zero in one of the parentheses?
x = -3 or x=2

3) x^2-6x+7=0         
Completing the Square:
Take half of b (-6), square it, and add/subtract that coefficient
x^2 - 6x + 9 - 9 + 7 = 0
(x -3)^2 - 2 = 0         Add 2 to both sides
(x-3)^2 = 2               Take the square root of both sides
(x-3) = √2 or -√2       Add 3 to both sides
x = 3 + √2 or 3 - √2