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# Solving Quadratic Equations in Disguise - Concept

###### Carl Horowitz

###### Carl Horowitz

**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

in here.

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.

Okay.

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

like this.

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

to 0.

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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###### Carl Horowitz

B.S. in Mathematics University of Michigan

He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his step-by-step explanations are easy to follow.

i love you you are the best, ive spent 3 hours trying to understand probability and this is making sense now finally”

BRIGHTSTORM IS A REVOLUTION !!!”

because of you i ve got a 100/100 in my test thanks”

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