Medal!An identity is a specific type of equation. Many equations are not identities, however, because an equation is not necessarily true for all values of the involved variables. Of the eight equations that follow, only four are identities. Label the equations that are identities as such and provide a counterexample for the equations that are not identities. a) ()()25525xxx−+=− b) ()22525xx+=+ c) 2xx= d) 42xx= e) 63xx= f) 22xyxy+=+ g) ()()2233abaabbab+−+=+ h) xyyx
Which ons are idetities
any ideas? for example, what do you get when you multiply \[(x+5)(x-5)\]?
The same answer?
we can check them one by one if you want that is really what you are supposed to do i guess
Yes please
is it clear that when you multipy \((x+5)(x-5)\) you get \(x^2-25\) ?
that makes it an identity, true for all \(x\)
Oh ok
now when you multiply \[(x+5)^2\] you get \[(x+5)(x+5)=x^2+5x+5x+25=x^2+10x+25\]
They don't gwt the same answer
since that is not \(x^2+25\) that is NOT an identity now you have to give an example, so pick a number, i don't care what you pick, pick anything (not too big)
2
ok now we do the part where they say "prove by counterexample it is not an identtity" if we put \(x=2\) in to \((x+5)^2\) we get \[(2+5)^2=7^2=49\] but if we put it in \(x+25\) we get \[2+25=27\] since \(49\neq 27\) that shows it is NOT an identity !
ready for the next one?
Yes
\[\sqrt{x^2}=|x|\] IS and identity if you square a negative number you get a positive one, take the square root and it will be the absolute value of what you started with for example \[\sqrt{(-5)^2}=\sqrt{25}=5\] and \[|-5|=5\] also so that one is an identity
\[\sqrt{x^4}=x^2\] is also an identity
Oh ok makesince
how we doing so far?
Good!
Wudnt e be a identity
looks like it, but it isn't that is because \(x^3\) can be negative, but \(\sqrt{x^6}\) cannot
Oh it cant
for example, if we pick any negative number (here comes the counterexample part) like \(-2\) then \[\sqrt{(-2)^6}=\sqrt{64}=8\] but \[(-2)^3=-8\] and since \[8\neq -8\] it is NOT an identity
Ok I think f is a identity tho
oh no you don't !!
\[\sqrt{x^2+y^2}\neq x+y\] pick any two numbers and try it!
lets make it easy and pick \(x=3,y=4\)
\[\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\] but \[3+4=7\]
O dang it
this is like pythagoras, two legs of a right triangle are greater in length than the hypotenuse
|dw:1411615088233:dw|
where we at, G?
X+y
we are at \[(a+b)(a^2-ab+b^2)=a^3+b^3\]right?
2^2+3^2=4+9=23 and 2+3=5 so no deal
well be careful, you left off the square root if you pick 2 and 3 it would be \[\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}\] and \[2+3=5\] but yeah, "no deal"
Oop wrong one ok next one was what u were doing
\[(a+b)(a^2-ab+b^2)=a^3+b^3\] that one IS an identity
that is how you factor the sum of two cubes if you want to prove it for yourself, multiply out on the left, combine like terms, and you will get \(a^3+b^2\)
and finally \[\frac{x+y}{x}=y\] is just ridiculous
pick any x and y you like, plug it in and you will see they are not equal
Lol
2,3
ok \[\frac{2+3}{2}\neq 3\] doh well that was exhausting hope it is more or less clear this is actually an
Wow that's so easy! Thank you a bunch
this is actually an intelligent introductory algebra question, one of the very few i see kudos to whoever wrote it
Its definitely more clear now
great yw
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