The product of 2 positive consecutive even integers is 48, what is the smaller number?
Two positive consecutive even integers, than mean 2,4,6,8,10,12.... Even numbers can be algebraically written as 2x, two consecutive, so the second number would be 2x+2
They are even so they have difference of 2: So, if one number be : x the other number will be: x + 2 So, x(x + 2) = 48 \[x^2 + 2x -48 = 0\] Can you factorize this equation??
Now let's compute an equation: The sum of those two numbers give 48; 2x+2+2x=48 And solve for x :)
@waterineyes x could be 5, and +2 would be 7 :s
@zepp it is product, not sum
guess and check for this one
I am assuming it as x...
@waterineyes yeah i can, will that give me the answer? And @zepp I'm so confused and this site is annoying me! will you message me and help me?
(2x+2)2x=48 then, thanks satellite :D
@DeaneNicole guess a number and see if it works
Yes you will surely get your answer just factorize that equation..
10 and 12 too big because \(10\times 12=120\) try 6 and 8
@satellite73 I'm trying to PASS my mastery test not fail it!!!
Yeah try 6 and 8..
but that is what you have to do. guess and see what works
@waterineyes I noticed that both methods work :D
there is no other way to do it, pick two consecutive even numbers (wisely) if their product is 48 you win, if not pick bigger or smaller numbers as needed. that is the only way to do this problem
@zepp ., do you know how both method works??
Um, doesn't work :U
if you want to write an equation you can write \[x(x+2)=48\] \[x^2+2x=48\] \[x^2+2x-48=0\] but then you are faced again with the original problem, finding two numbers that multiply to -48 and add to 2 it is the same problem exactly, only now you have an equation and you have made it harder
Do you know or not??
i agree with @satellite73 there is no other method other than guess n checking.. factorizing also u need to guess and put the same effort
Actually it works (2x+2)(2x)=48 4x^2+4x-48=0 Quadratic: x=3 & -4 2x=2*3=6 2x+2=6+2=8
@ganeshie8 thank you! i have argued this point ad nasueum and keep hearing "but you have to solve an equation" without realizing that solving the equation is identical to solving the original problem
See if you know want this then you can go with quadratic formula, there is no need of guessing there for factors...
in place of know there will come don't...
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