Algebraic Manipulation - IIT JEE Video Lecture

What I want to do
in this video is
something a little
unconventional.
Because I tried this problem
before I started this video,
and I kept getting
an answer that
isn't one of these
answers over here.
So assuming that the exam
writers did this properly--
and this is a serious exam.
It's to get into a very
prestigious set of universities
in India, so I'm
assuming they didn't
make any major mistakes here.
I want you to tell me where
I'm making the mistake.
So let's just try
to do the problem.
Let p and q be real numbers
such that p does not equal 0,
p to the third does not
equal q, and p to the third
also does not equal negative q.
If alpha and beta are
nonzero complex numbers
satisfying alpha plus beta
is equal to negative p
and alpha cubed plus
beta cubed is equal to q,
then a quadratic equation
having alpha/beta and beta/alpha
as its roots is.
Did I just call it b up here?
It's beta.
So what could that quadratic be?
Well, these are the
two roots, so that's
pretty straightforward.
We could say that the quadratic,
a possible quadratic here,
would be x minus alpha/beta
times x minus beta/alpha
is equal to 0.
Clearly, the roots of
this equation right
here are alpha/beta
and beta/alpha.
Either one of those would
satisfy this equation.
Now let's multiply this
out. x times x is x squared.
x times negative beta/alpha
is negative beta/alpha x.
Then you have
negative alpha/beta
x, so negative alpha/beta x.
And then you have
negative alpha/beta
times negative beta/alpha.
The alpha's cancel out.
The beta's cancel out.
The negatives cancel out.
And you get plus
1 is equal to 0.
And then let's see, we
might be able to simplify
this a little bit.
If we multiplied everything
times alpha beta,
what will happen?
So I'm just going to
multiply both sides
of this equation
times alpha beta.
I don't like having
things in denominators.
Clearly, the right-hand
side is still going to be 0.
The left-hand side now
becomes alpha beta x squared.
Alpha beta times beta/alpha,
the alphas cancel out.
You have a beta times a beta.
So it's negative beta squared
x minus-- the betas cancel out.
And now you have an alpha times
an alpha minus alpha squared x.
And now you have plus
alpha beta is equal to 0.
And then we can combine
these terms over here.
So we have alpha beta
x squared minus-- you
have an alpha squared plus a
beta squared x-- all right,
I just took the
negatives out of it--
plus alpha beta is equal to 0.
All right.
So everything I did is
just kind of-- I just said,
OK, these are the roots.
Let me just multiply it
out, and I got something
in terms of alpha and beta.
Now when you look at
it, their choices,
clearly they want answers
in terms of p's and q's.
So let's see if we can
convert this into p's and q's.
So a good place to start, they
tell us that alpha plus beta is
equal to negative p, or negative
p is equal to alpha plus beta.
And so if we squared that,
what I want to do is somehow
relate it to q as well.
And so maybe if we eventually
take it to the third power,
we can get some of these terms
over here and relate it to q.
And also, when we
square it, I think
we'll be able to relate some
of these things to it as well.
So let's square both
sides of this equation.
So you have negative
p squared, which,
because the negative
canceled out,
this is the same thing
as negative 1 times p
squared, which is negative
1 squared times p squared,
which is just p squared,
is equal to this expression
over here squared,
which is just alpha
squared plus 2 alpha
beta plus beta squared.
Fair enough.
And already we have
something interesting
showing up over here.
We can express alpha
squared plus beta
squared in terms of p
squared and alpha beta.
And then we simplify
things a little bit.
Well, we'll just keep going.
Let's just take this
to the third power.
So if we have negative p
to the third power, which
is the same thing as the
negative of p to the third,
if you multiply this
thing times negative p,
you're going to get the
negative sign p to the third.
And this is equal to-- we
could just multiply again,
do the distributive property.
Or if you kind of remember
Pascal's triangles,
you remember the coefficients
on it, or the binomial theorem,
this would be a
faster way to do it.
I'll do a Pascal's
triangle really quick here.
So you have 1.
1, 1 for binomial.
1, 2, 1, when you take the
square of a binomial, 1, 2, 1.
And then when you take
that to the third, when
you go to the third
power, it's 1, 3, 3, 1.
Bunch of videos on
this, especially
in the videos on the
binomial theorem.
So this to the third
power, this thing
right here to the
third power, is
going to be alpha
to the third plus
3 alpha squared beta
plus 3 alpha beta squared
plus beta to the third.
I just decreased the exponent
on the alpha each time
and started increasing the
exponent on the beta each time.
Or you could just
multiply this times this,
and you should also get this.
And now this is
interesting because this
is expressing everything
in terms of-- let's see
if we can simplify
this a little bit.
This is equal to--
so let me write
this-- we have negative
p to the third, which
is equal to negative
p to the third,
just a difference
of the parentheses,
is equal to-- we have this
alpha cubed plus the beta cubed.
That's those two terms
right over there.
And I separate those out
because those are our q.
Those are our q.
And then I have--
let's see, over here I
can factor out a 3 alpha
beta, so plus 3 alpha beta.
And what do I get?
If I factor out a
3 alpha beta here,
I'm just left with an
alpha on this term.
And if I factor out
a 3 alpha beta here,
I'm just left with a beta.
I am just left with a beta.
So what does this simplify to?
So this is going to simplify
to-- this right here
is equal to our q.
So this is equal to
q plus 3 alpha beta.
And what's alpha plus beta?
Well, they already told us
what alpha plus beta is.
Alpha plus beta is negative p.
It is negative p.
So let's see what
we can do over here.
And this right here
is the same thing
as negative p to the third.
So here we can solve
for alpha beta,
and so we'll have alpha beta
in terms of p's and q's.
So that'll be good.
So we can substitute the alpha
beta's with whatever we get
in terms of p's and q's.
And then this thing we can
express in terms of a p
and an alpha beta, which means
we can express it in terms
of p's and q's.
So let me do what
I'm talking about.
Let me do it I'm talking about.
So let's just solve, first
of all, for alpha beta.
So if we solve for
alpha beta, first we
want to subtract q from
both sides of this equation.
So you have negative
p to the third minus
q is equal to-- we have this
negative sign over here,
so it's equal to
negative 3 alpha beta p.
And I want to solve
for alpha beta, so let
me divide both
sides by negative 3.
And so we get, on
the right-hand side,
we get alpha beta is equal
to, if you divide-- so
let me just write it
out because I don't
want to make a careless mistake.
So you have negative p to the
third minus q over negative 3p.
Now if we multiply the
numerator and denominator
by a negative sign, we get p
to the third plus q over 3p
is equal to alpha beta.
So we can now express this
and this term in terms
of p's and q's.
Now what about this
one right here?
So what is alpha squared
plus beta squared?
And we see right up
here, alpha squared,
that's alpha squared
plus beta squared.
So if we subtract 2 alpha
beta from both sides
of this equation,
we get alpha squared
plus beta-- let me do this
in a different color-- we get
alpha squared plus
beta squared is going
to be equal to p squared
minus 2 alpha beta.
All right?
I just subtracted 2 alpha
beta from both sides.
And so this is going to be
equal to p squared minus 2 times
this thing, minus 2 times p
to the third plus q over 3p.
And what is this equal to?
This is equal to-- let's put
on a common denominator--
p squared over-- this is the
same thing as 3-- my cell
phone rang.
I had to pause it, sorry.
So this is equal to
3p to the squared--
if I want to put it
over 3p-- is equal to 3p
to the third over 3p.
Let me make it clear.
This term right here
is this term over here
if you were to simplify it.
The 3's cancel out.
You just have p squared.
And then you have
minus-- I'm just
going to distribute the negative
2-- minus 2p to the third minus
2q, all of that over 3p.
So it is equal to 3p
cubed minus 2p cubed
is equal to p cubed minus
2q, all of that over 3p.
So now we can write this
entire expression in terms
of p's and q's.
This first term over
here, alpha beta, this
is p cubed plus
q over 3p, and we
want to multiply
that times x squared.
And then you have minus alpha
squared plus beta squared.
We just figured out that
alpha squared plus beta
squared is this thing over here.
So minus-- I'll do it
in a different color,
I'll do it in green--
minus this thing,
p to the third minus 2q over 3p.
And that's going to be
multiplied by an x, and then
plus alpha beta again.
So alpha beta is p to
the third plus q over 3p.
Now, when we look at any
of the choices up here,
it looks simplified.
We don't see any 3p's
in the denominator.
So let's see if we
can simplify that.
Oh, and we don't want to
lose-- we had an equals 0 here
the whole time.
So this is equals 0.
So when you multiply both
sides of this equation by 3p,
what do you get?
This first term becomes p to the
third plus q times x squared.
The second term becomes minus
p to the third minus 2q x.
And then this last term becomes
plus p to the third plus q,
and this is going
to be equal to 0.
Now, let's see how this
compares to the choices.
And I think I actually did
not make my careless mistake
in this video.
I actually got the right answer.
Let me see.
It shows you, when
you do it carefully,
you're less likely
to make errors.
OK.
So let's see.
How does this match up
to what I have over here?
Well, this looks exactly.
I'm very excited because
I tried this twice
before actually
making the video,
and I kept making
a careless mistake.
But this time I didn't
make the careless mistake,
and I got B, p to the third
plus q x squared minus p
to the third minus 2q x
plus p to the third plus q.
So that right there, that
right there is our answer.
So I take back what I said
at the beginning of video.
I didn't need your help.
But I guess just the
process of explaining
it was the help that I needed.
Anyway, hopefully you
found that useful.