SOLUTION: the largest of three consecutive numbers is n. the square of this number exceeds the sum of the other two numbers by 38. find the three numbers.
Algebra.Com
Question 800497: the largest of three consecutive numbers is n. the square of this number exceeds the sum of the other two numbers by 38. find the three numbers.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
If n is the largest
then( n-1) & (n-2) are the preceding numbers
n^2=(n-1)+(n-2)+38
n^2=2n-3+38
n^2=2n+35
n^2-2n-35=0
n^2-7n+5n-35=0
n(n-7)+5(n-7)=0
(n-7)(n+5)=0
n= 7 OR -5
7,6,5
OR
-5,-6,-7
RELATED QUESTIONS
The largest of the three consecutive positive numbers is n. The square of this number... (answered by josgarithmetic)
The sum of the two smaller of the three consecutive numbers exceeds the largest by 52.... (answered by Boreal,lwsshak3)
Given that n is the first of three consecutive numbers and the sum of three numbers... (answered by ankor@dixie-net.com)
Given that n is the first of three consecutive numbers and the sum of three numbers... (answered by greenestamps)
the sum of two numbers is 38. three times the smaller number exceeds the larger number by (answered by nerdybill,biancafranco86)
the sum of two numbers is 38. three times the smaller number exceeds the larger number by (answered by 55305)
1.the sum of two is 5 times their difference.if one exceeds the other by 7,what are the... (answered by thinkingeye)
The sum of two numbers is 38. Four times the smaller exceeds three times the larger... (answered by flame8855)
Find three consecutive positive odd numbers such that the sum of the square of the two... (answered by macston)