Question 1014284
{{{n}}}= the smaller of the two odd integers
{{{n+2}}}= the next odd integer
{{{n(n+2)}}}= the product of those two consecutive odd integers
{{{n+(n+2)}}}= the sum of those two consecutive odd integers\
We translate "their product is 35 more than 8 times their sum" as
{{{n(n+2)=8(n+(n+2))+35}}} .
That is the equation we have to solve, to find {{{n}}} (and {{{n+2}}} ).
 
If we find a solution that is even, or negative, or not an integer at all, we discard it.
If we find a solution that is an odd integer, that is the answer to the problem.
If we do not find a solution that is an odd integer, the problem has no solution.
 
{{{n(n+2)=8(n+(n+2))+35}}}
{{{n^2+2n=8(n+n+2)+35}}}
{{{n^2+2n=8(n+n+2)+35}}}
{{{n^2+2n=8(2n+2)+35}}}
{{{n^2+2n=16n+16+35}}}
{{{n^2+2n-16n=16+35}}}
{{{n^2-14n=51}}}
That is a quadratic equation that can be solved 3 ways.
BY COMPLETING THE SQUARE:
{{{n^2-14n=51}}}
{{{n^2-14n+49=51+49}}}
{{{(n-7)^2=100}}}--->{{{system(n-7=10,"or",n-7=-10)}}}--->{{{system(n=10+7=17,"or",n=-10+7=-3)}}}--->{{{highlight(system(n=17,n+2=19))}}}
BY FACTORING:
{{{n^2-14n=51}}}
{{{n^2-14n-51=0}}}
{{{(n+3)(n-17)=0}}}--->{{{system(n=17,"or",n=-3)}}}--->{{{highlight(system(n=17,n+2=19))}}}
BY USING THE QUADRATIC FORMULA:
{{{n^2-14n=51}}}
{{{n^2-14n-51=0}}}
The quadratic formula, to find the solution to {{{ax^2+bx+c=0}}} is
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} .
In the case of {{{n^2-14n-51=0}}} , {{{system(a=1,b=-14,c=-51)}}} , and
{{{x=(-(-14) +- sqrt((-14)^2-4*1*(-51)))/(2*1)=(14 +- sqrt(196+204))/2=(14 +- sqrt(400))/2}}}--->{{{system(x=(14+20)/2=34/2=17,"or",x=(14-20)/2=(-6)/2=-3)}}}--->{{{highlight(system(n=17,n+2=19))}}}