SOLUTION:
Wondering if I'm right...
Find three consecutive odd integers such that the square of the third integer plus the product of the other two integers is 268.
(x+4)^2+x(x+2)=268
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION:
Wondering if I'm right...
Find three consecutive odd integers such that the square of the third integer plus the product of the other two integers is 268.
(x+4)^2+x(x+2)=268
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Question 1050894:
Wondering if I'm right...
Find three consecutive odd integers such that the square of the third integer plus the product of the other two integers is 268.
(x+4)^2+x(x+2)=268
x^2+8x+16+x^2+2x=268
2x^2+10x+16=268
x^2+5x+8=134
x^2+5x-126=0
(x+9)(x+14)=0
x=9,x=-14
ingegers=9,11,13
Thank you Found 2 solutions by ewatrrr, MathTherapy:Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! (x+4)^2+x(x+2)=268
x^2+8x+16+x^2+2x=268
2x^2+10x+16=268
x^2+5x+8=134
x^2+5x-126=0
(x-9)(x+14)=0 Note: Just the sign here
x=9, x=-14 (discard)
three consecutive odd integers = 9,11,13
checking
169 + 99 = 268 CHECKS!
Good work.
You can put this solution on YOUR website!
Wondering if I'm right...
Find three consecutive odd integers such that the square of the third integer plus the product of the other two integers is 268.
(x+4)^2+x(x+2)=268
x^2+8x+16+x^2+2x=268
2x^2+10x+16=268
x^2+5x+8=134
x^2+5x-126=0
(x+9)(x+14)=0
x=9,x=-14
ingegers=9,11,13
Thank you
Are you right? Just check it!
Yes, it is, so you are correct! Good job!!
However, when you factored the trinomial: , you should've gotten: (x - 9)(x + 14) = 0, not (x + 9)(x + 14) = 0.
This would've led to the 1st or smallest integer being - 9 (odd), or - 14 (even).
