level one here we have this linear equation x+1=0 we can just move the 1 to the other
side and we get x is equal to negative 1 and then we are done good level 2.
so now we have a quadratic equation x^2+x+1=0. well let's go ahead and do this by completing a
square first move the 1 to the other side so we will have x squared plus x and then that's equal
to negative 1. and for this part right here we are going to find the magic number so we can complete
the square and to do so we first make sure that we have a 1 in front of x squared which we do
then we see the coefficient of x which is 1 and we will have to take half of that and then square
work this out we will get 1 over 4. so this is the magic number now we have to add on both sides here
and here so the left hand side is going to give us x plus one half squared and then on the right hand
side this is going to give us negative three over 4. now we can just take the square roots on both
sides cancel this out and then put on the plus or minus on the left-hand side we get x plus one-half
but let's prove the one-half to the right side so x will be equal to negative 1 over 2 plus or minus
square root of 4 is just a 2 on the bottom and then we have square root of negative which is
going to be the i and then the 3 right still itself the square root so ladies and gentlemen
this right here is it great level three and now for this cubic equation x^3+x^2+x+1=0
we don't have to use the cubic formula we can actually just do this by factoring check this out
i'm going to do it by grouping for these two terms we can factor out the x square and then we will
just get x plus one and for these two terms we can just factor out like a one like this
and then we still have that x plus one and the nice thing is that the x plus one is in common
so we can factor that out like this and then we will have that x squared plus one left and
this is still equal to zero now put this to be zero and we also have to put this to be zero
so we see that right here just move the one to the other side x will be equal to negative
one and right here we can move the one to the other side and take the square roots on both
sides and don't forget the plus or minus right here we can get x is equal to plus or minus i
so we have a total of three answers negative one and the i and also the negative i
nice level four and now we have this quartic equation what do we even do the quartic formula
no don't do that check this out we are going to divide everybody by x^2 and you will see
that this will work out very nicely split it and divide them individually we get x squared plus
x plus 1 plus 1 over x plus 1 over x squared and that's equal to 0. now focus on the x squared plus
1 over x squared because in fact we can complete a square for them right here i'm going to write down
we have x squared plus 1 over x and then square this is the sum of two squares we can complete
a square by adding a middle term and the middle term is the following first we have the x squared
and then this right here is that 1 over x and then square if we add 2 times this and that
which is actually just a number 2 right this right here will give us the perfect square namely x plus
1 over x and then square very nice now of course if we added this term we will have to subtract it
and this is technically just what two right it's just a number two so we just have the minus two at
the end like so so now for this and that i'm going to replace it with this so we have parentheses
x plus one over x and then square and let me just put a minus two all the way at the end right here
and then we have this x plus one over x as well very nice huh so let's go ahead and add x plus one
over x and then of course we still have that plus one and minus two and that's equal to zero in fact
we have a quadratic equation in terms of x plus 1 over x so now let's just do a u substitution let u
equal to x plus 1 over x because this way we can see that this is u squared plus u and then this
is just minus one and that's equal to zero and because earlier we use the method of completing
a square so right now let's solve this by using the quadratic formula so u is equal to negative 1
plus or minus square root of 1 squared minus 4 times this and that which is going to be plus 4
so to get this 5 and then over 2 times 1. cool you know what's cooler this right here when we take
the positive negative one plus square root of five over two this right here is just the reciprocal of
the golden ratio and when we take the negative we will get negative 1 minus square root of 5
over 2 and this right here is just a negative golden ratio all right now we know that
x plus 1 over x is equal to this what is equal to that and we'll just solve them separately
and then right here i'm just going to multiply everybody by x and we will see this is x squared
plus one and then that's equal to this i'm going to move this to the other side so it becomes
negative and then i will just keep the reciprocal of the golden ratio and then x and then the
plus 1 here and that's equal to 0. and to solve this let's just use the quadratic formula again
x is equal to negative b which is going to give us just positive and then plus or minus square root
b squared well that will give us positive but i'm just going to write yeah i'm just going to write
it down like this and then minus 4 times 1 times 1 which is just minus 4 and then all over 2 times 1.
so from here we have two answers and then we will also have the negative golden ratio part
and then right here let's just multiply everybody by x again so we have the x squared and then
plus phi times x and then plus one and that's equal to zero here's the quadratic formula again
x is equal to negative v plus or minus square root and then we have the phi square and then
minus four times this and that which is minus four over two excellent level five finally we have
this quintic equation and of course i'll show you guys the quintic formula which says x is equal to
just kidding unfortunately there's no such a formula and in fact for this equation we can
solve it by factoring by grouping the first three terms and then the last three terms and we can go
from there but we did that earlier so should we get something else let's actually multiply
this equation by x minus one when we multiply the left-hand side we'll just get x to the sixth power
minus one and then the right-hand side will still give us 0. however notice that this equation has a
total of 6 solutions but originally we only have 5. and the reason is because we multiply out an
extra factor so just how to indicate that look at this but x cannot be equal to one all right
and now what's so good about solving this well we can move this to the other side so we just have to
look at x to the sixth power it's equal to one in fact we're just trying to find all the six roots
of one but we are not going to count x equal to one as a solution and to do so we can utilize
complex numbers take one to the complex world so i'll put that down right here for you guys
one is right here and then write this in the polar form we need two things the first thing
is the distance from the origin to here which is just one and then the next thing is the angle
but looking at this the angle is zero however we can keep rotating two pi over
the other way two pi right so we just have to add two um pi where an is an integer
so in fact for the one we can just look at this right here x to the sixth power
is equal to r is one and then we have the e and then i times theta so i times two pi like this now
we want to get the x right so let's just take both sides to the 1 over 6 power like so and
this end that will cancel so we are going to get x being equal to we can divide reduce and
all that stuff so this is going to be e and then we will just have i and then m pi over three but
what the n's though well keep this in mind when we want to find the sixth root of one all we have
to do is plug in n is equal to zero one two three four five and that's it if you continue they will
just give you duplicate the answers from here so don't do that and right here notice that we are
not going to use zero because if so we will get e to the zero power and that's one and we say x
is not equal to one so now we just have to plug in this right here so ladies and gentlemen x is
equal to plugging y into n we will get e to the pi over 3 i and of course you can use the euler's
formula to write the expression in the standard form but i'm just going to leave it like this
and then next we will have e to the 2 pi over 3 i and then e to the 3 pi over 3 which is just pi and
then i which we is actually just negative 1 and then e to the 4 pi over 3 i and lastly e to the 5
pi over 3 i all right very very cool here are the five solutions to the original equation awesome