• December 23rd 2011, 03:20 PM
Satnam
find the sum of 4a^3-7a-5 and -8a^2-2a+7 IS THE ANSWER

HOW TO do this The sum of four consecutive integers is at least 114. Find the smallest possible values for these numbers.
• December 23rd 2011, 03:28 PM
skeeter
Quote:

Originally Posted by Satnam
find the sum of 4a^3-7a-5 and -8a^2-2a+7 IS THE ANSWER
$(4a^3 - 7a - 5) + (-8a^2 - 2a + 7) = 4a^3 - 8a^2 - 9a + 2$
$n + (n+1) + (n+2) + (n+3) \ge 114$ solve for n