The product of two consecutive even integers is 120. Find the two integers.
you know what does consecutive mean ?
Yes, I'm just too lazy to think right now lol
10*12
lol, ok and they are even so, if smaller integer is x the larger integer will be x+2 do you get this ?
Thank you beautiful!! @Shay17 :)
yes, I get it.
yw
u still wanna know how to actually grt those ?
*get
My online work says I need two sets of numbers. a higher and lower set.
Yes, please because I dont think I'm completely done lol :/
thats right. so, product is 120 x(x+2) = 120 got this ? thats a quadratic and i thinky ou know how to solve quadratics :)
ehh, do I reaaaally have to think right now? lol
if you want to learn and practice, yes. anyways i would not give you the other set. i can make you find it out by yourself if you are willing to think :)
ugh, ok lol
\(x^2+2x =120 \\ x^2+2x-120 =0\) know how to solve this quadratic?
uhh, yeah? lol
good then tell me the 2 values of x you get.
-12 & 10 (Not gonna lie, I cheated, buuuuut I got the answers)
ok so, x and x+2 when x= 10, x+2 =12, {10,12} when x= -12, x+2 = -10, {-10,-12}
Ooooooh, I get it. Ok.
cool :)
Sooooo, what do I put for the different sets?
i just gave you 2 sets! {10,12} {-10,-12} thats it!
higher one is {10,12}
Oooooooh ! lol ok, thank you! :)
\[(x)(x+2)=120\] \(x\) is your 1st even integer; \(x+2\) is your 2nd even integer. \[x^2 + 2x = 120\] \[x^2 + 2x + 1 = 120 + 1\] \[(x+1)(x+1) = 121\] Now you see that the same number \(x+1\) multiplied by itself is 121. So, what is the square root of 121? It is 11. Since you know that \(x+1 = 11\), you know that \(x=10\), which is your first even integer. Now, the 2nd integer is \(x+2 = 10+2 = 12\). Just so you know how to sold the quadratic without cheating. :)
solve*
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