SOLUTION: Tim is thinking of three consecutive positive numbers. If he multiplies the first with the third and then add the second, the result is 41. Let x be the smallest number.
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-> SOLUTION: Tim is thinking of three consecutive positive numbers. If he multiplies the first with the third and then add the second, the result is 41. Let x be the smallest number.
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Question 92992: Tim is thinking of three consecutive positive numbers. If he multiplies the first with the third and then add the second, the result is 41. Let x be the smallest number. Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=the smallest integer
then x+1 is the next largest
and x+2 is the largest
x(x+2)=first multiplied by the third and this equals x^2+2x.
Now if he adds the second, he would have:
x^2+2x+x+1 and this turns out to equal 41. So our equation is:
x^2+2x+x+1=41 subtract 41 from both sides
x^2+2x+x+1-41=41-41 collect like terms
x^2+3x-40=0 quadratic in standard form and it can be factorsd:
(x+8)(x-5)=0
x=-8-----------------NO!!!! MUST BE POSITIVE
x=5----------------------SMALLEST INTEGER
x+1=5+1=6----------------NEXT LARGEST INTEGER
x+2=5+2=7 --------------LARGEST INTEGER
CK
Multiply 1st by 3rd: 5*7=35
Then add the second: 35+6=41
41=41
Hope this helps----ptaylor
