Asked by b
Are the two solutions to x (2x+1) = 1 found by setting x = 1 and 2x + 1 = 1? Explain your reasoning. What are the solutions to this equation? Show how you arrived at your answer. Create an equation for your classmates to solve that requires factoring.
Answers
Answered by
Steve
nope. Just because two numbers multiply together to make 1, you still know nothing about the numbers.
If nothing else, try your idea.
If x=1, 2x+1=3, and 1*3≠1
However, if two numbers multiply to be zero, then one or the other must be zero. So, let's set things equal to xero, rather than 1.
x(2x+1) - 1 = 0
2x^2 + x - 1 = 0
you can factor that to get
(2x-1)(x+1) = 0
Now you have two possibilities:
2x-1 = 0
x+1 = 0
So, that gives you two solutions for x:
x = 1/2
x = -1
To verify that they satisfy the original equation, plug them in:
(1/2)(1+1) = 1
(-1)(-1) = 1
So we are good.
If nothing else, try your idea.
If x=1, 2x+1=3, and 1*3≠1
However, if two numbers multiply to be zero, then one or the other must be zero. So, let's set things equal to xero, rather than 1.
x(2x+1) - 1 = 0
2x^2 + x - 1 = 0
you can factor that to get
(2x-1)(x+1) = 0
Now you have two possibilities:
2x-1 = 0
x+1 = 0
So, that gives you two solutions for x:
x = 1/2
x = -1
To verify that they satisfy the original equation, plug them in:
(1/2)(1+1) = 1
(-1)(-1) = 1
So we are good.
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