When five times the square of certain integer is divided by four more than three times the same number, the result is -10. What is the integer?
Answers
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Let’s call the lowest of these integers n.
Since all three of these numbers have to be odd, they will increase arithmetically by two. So the next numbers are n + 2 and n + 4.
The first half of this question asks us to find five times the square of the middle integer. We can pretty easily do this:
5(n+2)2=5(n2+4n+4)=5n2+20n+205(n+2)2=5(n2+4n+4)=5n2+20n+20
We will keep this is mind for now.
The second half of the questions asks us to find the number that exceeds the product of the other two numbers by 488. This just means the number 488 bigger than the product of the other numbers.
n(n+4)+488=n2+4n+488n(n+4)+488=n2+4n+488
The two equations we found can now be set equal to solve for n.
n2+4n+488=5n2+20n+20n2+4n+488=5n2+20n+20
4n2+16n−468=04n2+16n−468=0
n2+4n−117=0n2+4n−117=0
(n+13)(n−9)=0(n+13)(n−9)=0
So n is either -13 or 9. Since the question stated that n must be positive, it must be 9. So, the other two numbers are 11 and 13, found by plugging n back into n + 2 and n + 4.
Let’s check our work.
5(11)2=9(13)+4885(11)2=9(13)+488
5(121)=117+4885(121)=117+488
605=605605=605
Looks good! So the answer is {9, 11, 13}