Question 1159985: The square of a number increased by 3 times the number is 28. Find the positive number.
*I think I'm supposed to start with x^2(x squared)+3(x+1)=28, and then turn it into x^2(x squared)+3x+3=28, but I'm not sure of how to continue(or if this is right), so any help is appreciated.
Thanks!
Found 3 solutions by MowMow, ikleyn, MathTherapy:
Answer by MowMow(42)   (Show Source):
You can put this solution on YOUR website! x = 'the number'
x^2 is the square of the number
+3x is three times the number
So x^2 + 3x = 28
x^2 + 3x - 28 = 0
Factored:
(x-4)(x+7) = 0
x-4 = 0
x = 4
and (x+7) = 0
x + 7 = 0
x = -7
The number could be either 4 or -7
Answer by ikleyn(52787)  (Show Source):
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
The square of a number increased by 3 times the number is 28. Find the positive number.
*I think I'm supposed to start with x^2(x squared)+3(x+1)=28, and then turn it into x^2(x squared)+3x+3=28, but I'm not sure of how to continue(or if this is right), so any help is appreciated.
Thanks!
NO, you're WRONG!
Where did you get 3(x + 1) from? It says "3 times the number." Isn't that 3x, with x being the number?
Therefore, shouldn't the equation be: x2 + 3x = 28, which then results in: x2 + 3x - 28 = 0?
Solve that for x and you'll get 2 values for x. Then, give your answer based on this request: "Find the positive number."
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