SOLUTION: How would I find the real-number solutions of the equation: x^3-3x=0
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Question 239736
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How would I find the real-number solutions of the equation: x^3-3x=0
Answer by
Theo(13342)
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x^3 - 3x = 0
you can factor out the x to get:
x * (x^2-3) = 0
Either x is 0 or (x^2-3) is 0
x = 0 is one of the solutions.
x^2 - 3 = 0
add 3 to each side to get:
x^2 = 3
take square root of each side to get:
x = +/- sqrt(3)
those are your solutions.
x = 0
x = sqrt(3)
x = -sqrt(3)
a graph of your equation looks like this:
+/- sqrt(3) is roughly +/- 1.7632 as shown on the graph.