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A retail business has determined that its net income, in ter
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16 May 2018, 08:45
16 May 2018, 08:45
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A retail business has determined that its net income, in terms of x, the number of items sold, is given by the expression \(x^2 + x - 380\).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The number of items that must be sold for the net income to be zero |
10 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: A retail business has determined that its net income, in ter
[#permalink]
Updated on: 17 May 2018, 00:27
Updated on: 17 May 2018, 00:27
3
I solved the problem using the quadratic formula, but I welcome other ways on how to do it faster.
a=1, b=1, c= -380
x= \(\frac{-1+-\sqrt{1^2-4(1)(-380)}}{2(1)}\)
x= \(\frac{-1+-\sqrt{1521}}{2}\)
x= \(\frac{-1+39}{2(1)}\) and x= \(\frac{-1-39}{2}\)
X= \(\frac{38}{2}\) and x= \(\frac{-40}{2}\)
x=19 and x=-20.
Since we are looking for a positive integer, we can disregard -20 and only look at x=19.
Quantity A = 19
Quantity B = 10
Quantity A is greater.
The answer is A.
a=1, b=1, c= -380
x= \(\frac{-1+-\sqrt{1^2-4(1)(-380)}}{2(1)}\)
x= \(\frac{-1+-\sqrt{1521}}{2}\)
x= \(\frac{-1+39}{2(1)}\) and x= \(\frac{-1-39}{2}\)
X= \(\frac{38}{2}\) and x= \(\frac{-40}{2}\)
x=19 and x=-20.
Since we are looking for a positive integer, we can disregard -20 and only look at x=19.
Quantity A = 19
Quantity B = 10
Quantity A is greater.
The answer is A.
Originally posted by PIneappleBoy2 on 16 May 2018, 20:20.
Last edited by PIneappleBoy2 on 17 May 2018, 00:27, edited 1 time in total.
Last edited by PIneappleBoy2 on 17 May 2018, 00:27, edited 1 time in total.
Re: A retail business has determined that its net income, in ter
[#permalink]
16 May 2018, 21:41
16 May 2018, 21:41
1
PIneappleBoy2 wrote:
I solved the problem using the quadratic formula, but I welcome other ways on how to do it faster.
a=1, b=1, c= -380
x= \(\frac{-1+-\sqrt{1^2-4(1)(-380)}}{2(1)}\)
x= \(\frac{-1+-\sqrt{1521}}{2}\)
x= \(\frac{-1+39}{2(1)}\) and x= \(\frac{-1-39}{2}\)
X= \(\frac{38}{2}\) and x= \(\frac{-40}{2}\)
x=14 and x=-20.
Since we are looking for a positive integer, we can disregard -20 and only look at x=14.
Quantity A = 14
Quantity B = 10
Quantity A is greater.
The answer is A.
a=1, b=1, c= -380
x= \(\frac{-1+-\sqrt{1^2-4(1)(-380)}}{2(1)}\)
x= \(\frac{-1+-\sqrt{1521}}{2}\)
x= \(\frac{-1+39}{2(1)}\) and x= \(\frac{-1-39}{2}\)
X= \(\frac{38}{2}\) and x= \(\frac{-40}{2}\)
x=14 and x=-20.
Since we are looking for a positive integer, we can disregard -20 and only look at x=14.
Quantity A = 14
Quantity B = 10
Quantity A is greater.
The answer is A.
38/2 is 19 BTW.
How do we know square root of 1521 is 39?
I think we can just consider about the symobl the number would have and conclude as positive vs negative. Correct me if I am wrong.
Also is that the only way to solve this problem?
