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Solving Quadratic Equations in Disguise - Concept
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
There are four methods to solving quadratic equations: factoring, completing the square, using square roots, and using the quadratic formula. Sometimes there are more complex quadratic equations including equations that have fractional exponents and negative exponents. To solve these types of problems, we either make a substitution for a term or factor out negative exponents.
We're used to solving quadratics where
we have a second degree statement.
Typically what we're looking at is something.
of the form AX squared plus BX plus or something like
X squared minus Y squared and again
we could have coefficients
What we're used to doing is either factoring
those or completing the square
or quadratic formula, some sort
of way of solving them.
But what I want to talk to you now about
is what I call quadratics in disguise.
And basically things that we can use those
same methods to solve that don't
necessarily have the same
appearance as these guys.
And the first one is something of the form
AX to the 2N plus BX to the N plus C
is equal to 0. It's very similar
to what we have up here.
But we're throwing in these
extra little exponents.
We can actually end up factoring those
just as we can something like this but
instead of just using X as our first
term, this is like X minus something
times X plus something, we could do
the same thing here but we have X
to the N plus or minus something, times
X to the N plus or minus something.
So we can take the same exact approach that
we have up here to solve something
Likewise, if we have X squared minus Y squared,
these don't have to necessarily
be single variables.
They could be, say, F of X squared
minus G of X squared.
The difference of two functions squared.
Again, this is factored like X plus Y times
X minus Y we can do the same thing here.
F of X minus G of F of X. And my equal
So basically look for certain patterns that
you recognize. We're used to dealing with these.
These are no problem. But we can take the same
approaches we use to solve them when we see a pattern
and something that at least looks similar
to something that we already know.
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