SOLUTION: The product of the first and third of three positive consecutive even integers is 150 more than the sum of all three integers. This is what I tried: x(x+4)=x+x+1+x+4

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: The product of the first and third of three positive consecutive even integers is 150 more than the sum of all three integers. This is what I tried: x(x+4)=x+x+1+x+4      Log On


   

Question 468086: The product of the first and third of three positive consecutive even integers is 150 more than the sum of all three integers.




This is what I tried:
x(x+4)=x+x+1+x+4+150
x^2+x-155 <--I was stuck at this point (Try using the Quadratic formula, didn't work.) Thank you for your help!

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Let x, (x+2) and (x+4) represent the three positive consecutive even integers 

Question states***

 x(x+4) = [x + highlight%28x%2B2%29 + (x+4)] + 150

Solving for x

 x^2 + 4x = 3x + 156

 x^2 + x  - 156 = 0

factoring

(x +13)(x-12) = 0

(x +13)= 0   x = -13  Extraneous solution

(x-12) = 0   x = 12

the three positive consecutive even integers are 12,14,16



CHECKING our Answer***

 12*16 = 192 = 42 + 150  = 192