Diesel Posted January 5, 2005 Posted January 5, 2005 I'm having a problem in this please i would really appreciate if someone can solve this for me with the formula and everything 1- Find the whole number such that four times the number subtracted from three times the square of the number makes 15.
matt grime Posted January 5, 2005 Posted January 5, 2005 Let n be an integer such that 3n^2 - 4n=15, find n. Can be done in two ways. one you can use the quadratic formula, but better, factor the left and right and conclude something using uniqueness of prime factorization
ed84c Posted January 5, 2005 Posted January 5, 2005 just work through the formula on paper, and you should get that (well I did any way :-S)
Callipygous Posted January 5, 2005 Posted January 5, 2005 I'm having a problem in this please i would really appreciate if someone can solve this for me with the formula and everything 1- Find the whole number such that four times the number subtracted from three times the square of the number makes 15. 3n2-4n=15 3n2-4n-15=0 n=(4(+/-)sqrt(16-4(3)(-15)))/6 n=(4(+/-)sqrt(16+180))/6 n=(4(+/-)14)/6 n=(2+7)/3 n=3 or n=(2-7)/3 n=-5/3
matt grime Posted January 5, 2005 Posted January 5, 2005 In all honesty it is far better to do it with integer factorizations: n(3n-4) = 3*5 though in this case you need to consider -3*-5 as an option too. I say this because this techique allows you to solve equations like: find all primes, p such that p+9=n^2 where n is an integer, and we do not even need to consider anything as high faluting as a formula.
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