In this lesson, we are solving equations quadratic
in form by using substitution.
The first equation is 
x to the 4th plus 3x squared minus 4 equals 0.
I want to re-write this equation in the form a times 
something
in parentheses squared plus b times the same 
thing
in parentheses plus c equals 0.
Well, in this case, I can write this as x squared,
squared plus 3 times x squared
in parentheses again, minus 4 equals 0.
We will do what we call a "u" substitution for the 
thing
in parentheses, which is x squared here,
and that gives me u squared plus 3u minus 4 
equals 0,
which factors into u plus 4 times u minus 1 equals 
0.
We set each factor equal to 0,
resulting in u equals negative 4 and u equals 1.
Remember we're solving for x, not for u,
so I will replace the u with x squared
in each of these equations.
x squared equals negative 4, x squared equals 1.
x squared equals negative 4 has no real solution.
x squared equals negative 1 has the solutions
of negative 1 and positive 1.
So our solution here is x equals negative 1 or 
positive 1.
Next is an example you can try on your own.
Solve the equation x to the 6th plus 7x cubed 
equals 8
using substitution.
The next equation is t to the 2/3 minus 2t
to the 1/3 minus 3 equals 0.
I can re-write this as
t to the 1/3 squared minus 2t to the 1/3...
minus 3 equals 0.
And letting u be equal to t to the 1/3 here,
I will have u squared minus 2u minus 3 equals 0,
factoring into u minus 3 times u plus 1 equals 0.
Setting each factor equal to 0 gives me u equals 3
and u equals negative 1.
This time we will replace u with t to the 1/3,
so I have t to the 1/3 equals 3,
and t to the 1/3 equals negative 1.
Cubing each side will give me, first of all, t equals 
27,
and then t equals negative 1.
So the solutions, t equals 27 and t equals negative 
1.
The next problem is x plus square root of x equals 
20.
As you notice, we have a 20 on the right-hand side
where I would preferto have a 0.
So I'm going to start by subtracting 20 from each 
side...
and then looking at x as being the square root
of x squared plus the square root of x minus 20 
equals 0.
Here I will let u equal the square root of x
and re-write my equation
as u squared plus u minus 20 equals 0.
This factors into u plus 5 times u minus 4 equals 0.
Setting each factor equal to 0, I will have u equals 
negative 5
and u equals 4 and then replacing the u with 
square root of x.
 Square root of x equals negative 5.
Square root of x equals 4.
Square root of x equals negative 5 does not have 
a real solution.
Square root of x equals 4 has the solution x equals 
16.
And we will move onto the next equation.
2x to the negative 2 plus 5x
to the negative 1 minus 12 equals 0.
Here, let's re-write a bit.
Beginning with 2 times 1 over x squared plus 5 
times 1
over x minus 12 equals 0.
And one step further, before we do a u substitution,
let's write this as 2 times 1 over x squared plus 5 
times 1
over x minus 12 equals 0.
Here u will be equal to 1 over x, and I can write my 
equation
as 2u squared plus 5u minus 12 equals 0,
factoring into 2u minus 3 times u plus 4 equals 0.
Setting each factor equal to 0 will give me u equals 
3/2
and u equals negative 4.
And then substituting 1 over x for u, I will have 1
over x equals 3/2, and 1 over x equals negative 4.
Simplifying gives me x equals 2/3 and x equals 
negative 1/4.
Okay. So our two solutions here.
One thing to keep in mind is -- the quadratic 
formula,
there may be occasions using substitution
where you will not be able to factor, and you will 
need
to use the quadratic formula.
