us vs 4 non-real math questions [Q1] sqrt(x)+sqrt(-x)=2 [Q2] ln(x)+ln(-x)=0 [Q3] e^x+e^(-x)=0 & [Q4] final boss
this equation has a real solution this equation also has a real solution
but what if we add them up and we get square root of x plus square root of negative x
and that's equal to 2. unfortunately this equation does not have any real solutions anymore
and let's talk about why first let's consider the left-hand side and i'm just going to call
that to be f of x and the deal is that we have to worry about the domain of this function
whenever we have a square root we will have to pay attention to the inside
we will have to first make sure that this x has to be greater than or equal to zero because we
can have a negative number in the square root because otherwise we will get complex numbers
and for the second one we will have to put this to be greater than or equal to 0 as well
but this right here is a negative x so right here dividing negative 1 on both sides we are going to
need x has to be less than or equal to zero but if you guys take a look right here combining
these two conditions the only number that can make this work is when x is equal to zero so
you know what in fact this function is just a point because we can only plug in zero into here
and we are only going to get the word head so if you ever feel like the need to graph the origin
go ahead and just use this function but now let me just tell you of course you guys saw the shirt
already we'll be talking about complex solutions for this and this is how you can do it so here we
go we see that we have a negative inside of the square root right here so i'm just going to keep
this as how he is but then for this i'm just going to put out an i so we'll just say plus i and then
square root of x and that's equal to 2. now let's go ahead and factor out the square root of x and
i'm going to write down parenthesis 1 plus i first and then square root of x and that's equal to 2.
and then i will divide this on both sides and then of course we can just go ahead and square
both sides two squared which is four and then one plus i squared we will have to do
one square which is one and then plus two times this and that which is two i and then lastly we
do i squared which is negative one and this cannot cancel and then we can also reduce this and that
so we will get two over i but i don't know about you guys but i don't like to
be on the bottom so let's multiply the top and bottom by and then we get this right here being
equal to negative one so only always see that x is equal to negative and then we have the two i
and now here's a question for you guys is this the only solution no but if you see the whole process
it only gives us this one right so we will have to go back to the original equation
the deal is that this equation is symmetric because here we have x for the input and here
we have negative dx for the input so that means if negative 2i is a solution that means plus d2i
will also be a solution so i will just include that and also x equals 2i and then we are done
okay same deal this equation has a real solution this equation also has a real solution but
how about when we add them up l and x plus ln of negative x is equal to zero
let me tell you no more no more real solutions and let's talk about
why so let's do what we did earlier let's consider the function f of x being equal to ln of x plus
ln of negative x and remember in order for us to get real numbers we will have to make sure the
input is bigger than zero so the first condition is we have to make x being greater than zero
we don't have the equal sign here and then pay attention to this right here we will actually need
negative x to be greater than zero as well meaning that x has to be less than zero but if you look at
these two conditions is it possible for us to find a number that's bigger than zero and less
than zero at the same time no impossible so this right here is even worse because earlier
we have at least one point right now we don't even have any point
so this is like really like the empty graph if you would like it's just like the x and y axis
but of course let's talk about how you can find the complex students
can we do what we did earlier i would put out the negative yeah because you can actually look
at this as ln of negative 1 plus ln of x and then you can actually work out the ln of negative 1.
but for this one you don't have to do that because the key is this plus sign let's
just go ahead and combine his algorithms so nice you'll see so together we will have ln
and we are going to put these two together inside and we are going to multiply them so it will be
x times negative x and of course this is still equal to zero now x times this we get negative
x squared and that's equal to zero so you can see that the inside actually always negative so yeah
no real solutions for here and we can actually solve this though we can just exponentiate both
sides so this means negative x squared has to be 1 divide negative 1 on both sides and then just
take the square roots on both sides as usual and of course don't forget the plus minus ladies and
gentlemen the solutions for this one is x is equal to plus or minus square root of negative 1 is i
yeah really cool huh all right another one first equation e to the x is equal to zero and let me
tell you no solution not even complex solutions all right and please do not say negative infinity
because negative infinity is not a number so this right here has absolutely no solutions
and then for the second one let me tell you this right here also no solution not even the complex
numbers either and please do not say positive infinity and just like what we did earlier what
if we add them up we're looking at this equation e to the x plus e to the negative x is equal to zero
let's talk about the domain this time like earlier let's consider the function f of x being equal to
e to the x plus e to the negative x in fact the domain for this one is so nice because there's no
restriction for what number you can use for x so i will just write this down real quick the domain
is actually just all real numbers so that's the symbol for that and in fact if you graph this
actually let me come back here and just show you guys the graph for e to the x which is of
course the exponential graph like so and then e to the negative dx is just the other direction
exponential decay cool and you can just kind of add the graphs together all right because
you can see that this is almost zero and add this so it's going to curve up and then likewise this
so you actually end up with something like this yeah it's not a parabola it's almost like a
function all right cool so you see that we do have a curve much better than earlier right
but this curve will never touch the x-axis that's why it has no
let me tell you it has no solution but for this one we do have complex solutions
for this one and this one they have absolutely no solutions even though it's the similar situation
you see the curve right here does not touch the x-axis this curve here does not touch the x-axis
and you may think that there's like some imaginary solutions but no not at all absolutely but for
this one magically yes so have a look let's find them all we have a lot so here is how we can do it
um let's move this to the other side so we have e to the x and that's equal to negative e to the
negative dx and right here let's get rid of the negative exponent by multiplying this equation
with e to the plus dx this times this will give us e to the 2x and then this times that will just
give us one because the exponent has start to be zero so we have negative one ah-ha how do we
deal with this still go ahead and take the ln on both sides this way this and that
cancel but now the question is how do we handle ln of negative one want to know a quick way to do it
the quick way is this is actually on my shirt already we can write negative 1 in the polar form
namely e to the i pi yeah so right here we are going to get 2x is equal to ln and then this right
here is e to the i pi i can prove to you guys that this is negative 1. you see that's the proof
but the truth is this is not the only way to get negative 1 because the pi is not the only
angle in the polar plane that will give you this complex number to end up with negative one so the
deal is that we look at this pi here plus two and pi and in fact you can do it now you can do later
up to you but perhaps i'll just do it now and if i do this now i can cancel that and i'm going to
just put an entire thing right here so ladies and gentlemen let's just divide the two on both sides
so ladies and gentlemen for this one we have infinitely many complex solutions even the
earlier these two equations they were not possible anyway check this out for this equation in fact we
have infinitely many students and i'll tell you x will be equal to pi over 2 and that's actually
the first angle that will make sine equals 1. but we have infinitely many of them because we
can start from here and just keep adding 2 pi so we can just say plus 2 um pi and now let's
put down n is an integer right here so wow we have infinitely many solutions for this equation
similarly for the second equation x is equal to 3 pi over 2 and then plus 2 and pi
where n is an integer okay so we also have infinitely many solutions for this
and again we are going to add them up 6x plus sine of negative x is equal to 2. you
think i'm going to do it for you no i'm actually going to leave this for you guys to think about