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.

Question

anonymous · Answer

Someone pleaee help.mee

mertsj · Answer

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 · Answer

I got x^2 -4x-12 =O

mertsj · Answer

$$x^2+6x+8=10x+20$$
$$x^2-4x-12=0$$
$$(x-6)(x+2)=0$$
$$x=6, x+2=8, x+4=10$$
$$x=-2, x+2=0, x+4=2$$

mysesshou · Answer

Good !
last line unused for answer because not all positive even integers

anonymous · Answer

Im sorry but I dont get it

mertsj · Answer

You got the right equation.  Factor it and solve.

mertsj · Answer

Can you factor it?

anonymous · Answer

I only got up to (x-6) (x+2) = O

mertsj · Answer

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 · Answer

You would get x^2 - 4x - 12 = O

mysesshou · Answer

x-6=0 
 x+2=0
solve for x first

mertsj · Answer

When you solve x-6=0 you get x^2-4x-12=0?????????

anonymous · Answer

Ur answer wud be x = 6 and x = -2

mertsj · Answer

Yes.  But -2 is not positive so discard it and then find the other two consecutive even integers if the first one is 6