to solve this equation i'm going to use
substitution the key to substitution is
the middle term so I'm going to let u
equal x to the minus 1 power so then u
squared would be X to the minus 1 power
squared however the power you multiply
negative 1 times 2 is negative 2 so X to
the minus 2 is equal to u squared that's
what we have here so making that
substitution I'll be left with u squared
minus u minus 1 equals 0 so you can see
this is a regular old quadratic turns
out though this one's not going to
factor so let's use the quadratic
formula a is 1 B is negative 1 and C is
negative 1 using the quadratic formula
notice here my variable is u so u equals
negative B plus or minus b squared minus
4ac all over 2a let's go ahead and plug
in so negative B negative negative 1
plus or minus the square root negative 1
squared minus 4 times a times C all over
2 times a is 1 then simplifying we'd be
left with 1 plus or minus the square
root negative 1 squared is 1 plus 4 all
over 2 and negative 4 times 1 times a
negative 1 is positive 4 or 1 plus or
minus the square root of 5 all over 2
okay so here we got two solutions for
you you could equal one minus the square
root of five over two or you could equal
1 plus the square root of five over two
two solutions for you but we're solving
for x let's remember x equals our X to
the negative 1 equals you so when we
make that stack substitution u equals x
to the negative one like that okay so
that should equal one minus square root
of five over two or again back
substitute
X to the minus one should equal 1 plus
square root of 5 over 2 now we have to
solve each one of these remember X to
the negative 1 power is the same thing
as 1 over X so we can write that as 1
over x equals 1 minus square root of 5
over 2 or 1 over X could also equal 1
plus square root of 5 over 2 okay at
this point we can cross multiply and
then solve for x but I'm gonna show you
a little bit of a trick whenever you
have a fraction equals a fraction you
can reciprocate both sides so x over 1
should equal 2 over 1 minus square root
of five also x over 1 could equal 2 over
1 plus square root of 5 that's just
reciprocating both sides now the
solutions in the back of the book
probably have the answers rationalize
you can't have radicals in the
denominator so to do that if you
remember what we're going to do here is
multiply numerator and denominator by
the conjugate here 1 plus square root of
5 same thing on this one the conjugate
though would be 1 minus square root of 5
okay so let's simplify these that then
says x equals 2 times 1 plus square root
of 5 all over when you multiply these
together 1 times 1 is 1 the middle terms
are going to cancel and the negative
square root of five times square root of
5 is minus 5 and then over here we'll be
left with 2 over 1 minus square root of
5 you know over 1 again minus Phi so we
could simplify these a little more 2
times 1 plus square root of 5 all over
here 1 minus 5 is negative 4 same thing
here 1 minus square root of Phi 2 times
1 minus square 2 5 over negative 4 and
then we can cancel it looks like for our
final answer we would have 1 plus square
root of 5 over negative 2 I tues would
cancel
then here we have 1 minus square root of
5 over 2 okay so cleaning this up a
little further then X would be equal to
negative one-half minus square root of 5
over 2 or X would be equal to one-half
whoops sorry that should be negative so
negative one-half plus square root of 5
over 2 so in the back of the book you
might see the two solutions expressed in
set notation negative one-half plus or
minus the square root of 5 over 2 so
there's my two solutions to the original
using u substitution
