Need algebra help with quadratic equations

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If I have 599 matchsticks to make my last two figures what would the answer be if my equation is 4+3(n-1) and 4+3(n-2)

Oct 3rd, 2015

Thank you for the opportunity to help you with your question!

Since I don't know what your last two figures are, I'll have to guess. If I'm wrong, let me know. I'm guessing that the figure is a rectangle with the short side equal to 4 + 3(n-1) matchsticks in length and the longer side equal to 4 + 3(n-2) matchsticks in length. The area of the rectangle would be A = (4 + 3(n-1)) *(4 + 3(n-2)), If I want to fill the rectangle using exactly 599 matchsticks, then I must have A = 599. Solving for n, I have

599 = (4 + 3n -3) * (4 + 3n - 6) = (1 + 3n) * (-2 + 3n) = (3n +1) * (3n - 2) = 9n^2 - 6n + 3n - 2 = 9n^2 -3n +2

or 9n^2 - 3n - 597 = 0. Using the quadratic formula with n in the of x, n ~ 8. You can then back substiture. Since n is not actually a whole number, I suspect my initial guess is incorrect. Clarify and I will correct.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 20th, 2015

4+3(n-1)=m to the n power


Sep 20th, 2015

use the same equation for the second part except it will be n-2

Sep 20th, 2015

It took 599 to make the last two in the squence

Sep 20th, 2015

The # 3 question is the one I am trying to work out

Sep 20th, 2015
img_0159_1_.jpg

The # 3 question is the one I am trying to work out

Sep 20th, 2015

I'll review the problem in about 5 minutes.

Sep 20th, 2015

The answer is n = 99. I'll send you a table in few minutes that explains why.

Sep 20th, 2015
Sep 20th, 2015

Best wishes.

Sep 20th, 2015

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Oct 3rd, 2015
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Oct 3rd, 2015
Jun 28th, 2017
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