Question

three consecutive positive integers are such that three times the middle integer exceeds the sum of other two by 15. Find the middle integer​

Answer 1

Answer:

let x be the first consecutive positive integer, then

x, x+1, x+2 are the three positive consecutive integers and p = x+1

now we are given,

3*(x+2)^2 = x^2 + (x+1)^2 + 67

calculate squares

3*(x^2+4x+4) = x^2 +(x^2+2x+1) +67

multiply and combine like terms

3x^2+12x+12 = 2x^2+2x+68

subtract 2x^2 from both sides of =

x^2 +12x +12 = 2x +68

subtract 2x +68 from both sides of =

x^2 +10x -56 = 0

fact this polynomial

(x-4)*(x+14) = 0

x is 4 or -14

we want a positive integer, therefore

x = 4 and

p = 4 +1 = 5

check our answer

3*(6^2) = 4^2 +5^2 +67

108 = 16 +25 +67

108 = 108