WEBVTT
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Good news, there's a class
of equations in which we
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are already experts.
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When f is a polynomial, the
equation f of x equals 0
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defines the roots
of the polynomial.
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We have studied in detail the
issue of finding these roots.
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Recall, for example,
the following fact
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for the quadratic
polynomial case.
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We know that there is something
there, the discriminant, which
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will tell us an awful lot about
the roots of this polynomial.
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Capital Delta-- by the way,
there's another beautiful Greek
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letter--
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is the discriminant,
b squared minus 4ac.
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We know already from
this proposition
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that we've seen and proven that
if Delta is strictly negative,
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there are no roots.
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And if delta is
positive, then there
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are roots of the
polynomial, and we
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know how to calculate them
from this quadratic formula
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that we will know till
the end of our lives.
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Now it's a fact that
in specific examples
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you sometimes have to
massage the given equation
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to turn it into polynomial form.
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A notable example of this
is when rational functions
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are involved.
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Here's a case in which you
want to solve an equation where
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the sum of two rational
functions equals 3.
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Remember, rational
function means
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polynomial over polynomial.
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Now the natural domain
of this equation
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is the whole real line, but take
away the points 1 and minus 2
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for obvious reasons.
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We multiply both sides of
the equation by the product,
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x minus 1, x plus 2.
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This will have the
effect of getting rid
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of the denominators
on the left-hand side,
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so, algebraically, it will
simplify our equation.
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In doing this
multiplication, by the way,
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we're not multiplying by
anything that can be 0,
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so we are preserving equivalence
because this term will not
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be 0 in the domain
as we've defined it.
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So after this multiplication,
we come upon this equation.
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Now we have to simplify it.
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Our mastery of arithmetic
allows us to do this with ease.
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And when the dust settles,
you have a quadratic equation
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for x.
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It turns out that it factors,
and, therefore, its roots are
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evident, minus 5 and plus 2.
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And the solution set of
our original equation
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is minus 5 and 2.
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Two points that
are in the domain.
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Now that should be right
because we haven't done anything
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to change equivalence,
but we're not androids,
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so we check that it works
in the original equation.
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How about polynomial equations
of degree greater than 2?
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Let me remind you that we
had a strategy for finding
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roots of a general polynomial.
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It was based upon writing
the root factorization
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of the polynomial,
that means writing it
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as a product of linear
factors, x minus r,
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times a polynomial that
has no further roots.
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And the way to obtain
such a root factorization
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involved order reduction and
the fun of polynomial division.
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Well, we won't go
back all over that,
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but let's look at the special
case of a pure n-th order
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equation, an equation like this,
a x to the n plus b equals 0.
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So a here is different from 0.
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And this equation is
equivalent to x to the power n
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equals minus b over a.
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Since we want to find x, we
are very tempted to simply take
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the n-th root.
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But remember that
the n-th root of y
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is only defined
when y is positive.
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Therefore, we have to be a
little more careful in solving
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the equation and what
happens next depends entirely
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upon the parity of n.
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Is n even, or is n odd?
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If n is even and if minus b
over a is strictly negative,
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then there are obviously no
solutions to our equation
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because an even power of x can
never be strictly negative.
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On the other hand, if
minus b over a is positive,
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then the n-th root
of that number
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is a solution, but
since n is even, also
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minus the n-th root
is the solution.
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So we have two
solutions in that case.
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In the remaining
case, when n is odd,
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there will always
be one solution,
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but it will be written
somewhat differently
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depending upon whether minus b
over a is positive or negative.
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When it's positive, it's
just directly the n-th root.
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When it's negative, we
have to write the solution
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as minus the n-th root of the
positive number, b over a.
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Now let me give you
an example in which we
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will use a frequently useful
technique called change
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of variables to simplify a given
equation that we want to solve.
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Suppose we have a polynomial
equation like the example you
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see in which there is
a square power of x
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and a fourth power of x,
but there's no cubic power
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and there's no x.
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This suggests a
certain useful change
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of variables that might
simplify the equation.
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We're going to introduce
a new variable.
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I'll call it capital
X, and it's going
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to be a stand-in for x squared.
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So that's our
change of variables.
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Why do we make this
change of variables?
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Well in the hope that when we
express our original equation
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with the new variable
it will be simplified.
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And you see this is the case
because the new equation,
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in terms of capital X, is
a straightforward quadratic
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equation, capital X squared
minus capital X minus 2
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equals 0.
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It's an equation we can
easily solve by factoring.
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We find that capital X
must be 2 or minus 1.
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Now we go back to the
original variable,
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capital X is little
x squared, so that
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means x squared has
to be 2 or minus 1.
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Now x squared can't be minus
1, of course, therefore,
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we retain only the possibility
that x squared equals 2.
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And this gives us two roots, x
equals plus and minus root 2.
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So the solution set is
the set of two points
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plus and minus root 2.
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And we check our
answer, of course.
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Let me remark that the
change of variables X
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equals x cubed is one that will
do very good work when we apply
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it to an equation such as x to
the sixth minus x cubed minus 2
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equals 0.
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It's pretty much the same idea.
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