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Question 1006087: The sum of the cube of a number and twelve times the number is equal to 7 times the square of the number. Find the number
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = the number.
the cube of the number becomes x^3
12 times the number becomes 12x
7 time the square of the number becomes 7x^2
you get:
x^3 + 12x = 7x^2
subtract 7x^2 from both sides of the equation to get:
x^3 - 7x^2 + 12x = 0
factor out an x to get:
x * (x^2 - 7x + 12) = 0
factor the quadratic to get:
x * (x-4) * (x-3) = 0
set each of the factors equal to 0 and solve for x to get:
x = 0
x = 4
x = 3
those are the values of x that makes the original equation true.
the original equation is:
x^3 + 12x = 7x^2
when x = 0, you get 0 = 0 which is true.
when x = 3, you get 63 = 63 which is true.
when x = 4, you get 112 = 112 which is true.
you have 3 possible values of x that satisfy the constraints of the problem.
they are x = 0,3,4
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