anonymous
  • anonymous
find three consecutive positive even integers such that the product of the second and third integera is twenty more than ten times the first integer. Someone please help.me.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Someone pleaee help.mee
Mertsj
  • Mertsj
x= first even integer x+2 = second even integer x+4= third even integer (x+2)(x+4) = product of the second and third integers \[(x+2)(x+4)=10x+20\]
anonymous
  • anonymous
I got x^2 -4x-12 =O

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Mertsj
  • Mertsj
\[x^2+6x+8=10x+20\] \[x^2-4x-12=0\] \[(x-6)(x+2)=0\] \[x=6, x+2=8, x+4=10\] \[x=-2, x+2=0, x+4=2\]
mysesshou
  • mysesshou
Good ! last line unused for answer because not all positive even integers
anonymous
  • anonymous
Im sorry but I dont get it
Mertsj
  • Mertsj
You got the right equation. Factor it and solve.
Mertsj
  • Mertsj
Can you factor it?
anonymous
  • anonymous
I only got up to (x-6) (x+2) = O
Mertsj
  • Mertsj
If you multiply two numbers and get 0 then one of those numbers has to be 0. So write x-6=0 or x+2=0
anonymous
  • anonymous
You would get x^2 - 4x - 12 = O
mysesshou
  • mysesshou
x-6=0 x+2=0 solve for x first
Mertsj
  • Mertsj
When you solve x-6=0 you get x^2-4x-12=0?????????
anonymous
  • anonymous
Ur answer wud be x = 6 and x = -2
Mertsj
  • Mertsj
Yes. But -2 is not positive so discard it and then find the other two consecutive even integers if the first one is 6

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