Question 916180:
1.Out of three consecutive positive integers,the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Please help me to solve this problem, which is based on quadratic equation.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 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
our answer checks
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