Posted by Christopher on .
When you use the quadratic formula (x= b+/=......) can you use that on equations like X^4+2X8?
If not, how would you algebraically find the roots of: X^4+2X8?
Thanks!

Math 
Damon,
no you may not.
I do not know any easy way to do what you wish.
If it were
x^4 + 2 x^2 8
that would be easy
let z = x^2
then
z^2 + 2z  8 = 0
z = 2 or z = 4
then x = +/ sqrt 2
or x = +/ 2i 
Math 
Count Iblis,
You can't use the quadratic formula for such an equation. But you solve third and fourth degree equations algebraically and the way that is done uses a lot of the techniques that arise in case of quadratic equations.
Solving a general fourth degree equation (one that doesn't have an obvious solution that you can find by attempting some trivial factorization) involves completing a square, just like in case of the derivation of the quadratic formula.
In the case at hand, you can write:
x^4 = 2x + 8
If we can make both sides a perfect square, we can extract the square root. But the right hand side is not a perfect square. The left hand side is x^4 = (x^2)^2, and that is obviously a perfect square. The trick is then to add a parameter y inside the square so that as a function of y, it is always a perfect square:
(x^2 + y)^2 = x^4 + 2 x^2 y + y^2
Then the equation you want to solve implies that this is equal to:
2 x^2 y  2 x + y^2 + 8
So, we have the equation:
(x^2 + y)^2 = 2 y x^2  2 x + y^2 + 8
Note that this equation is equivalent to the eqution you want to solve, no matter what you chose for y. So, the solution for x is not affected by y.
Moreover, no matter what you chose for y, the left hand side is always a perfect square. What we then can do is to chose a special value for y such that the right hand side becomes a perfect square.
Here what you've learned from solving quadratic equations can be used. The term inside the root of the quadratic equation (b^2  4 a c) is called the discriminant, if this is zero then the quadratic expression is a perfect square. So, we need to calculate the discriminant, equate that to zero and solve for y.
We have:
a = 2 y
b = 2
c = y^2 +8
So:
b^2  4 a c =
4  8 y (y^2 + 8) =
8 y^3 64 y + 4
So, we first need to solve the equation:
y^3 + 8 y  1/2 = 0
Let's skip this step and assume that we have the solution to this equation. Then you just plug that into the equation:
(x^2 + y)^2 = 2 y x^2  2 x + y^2 + 8
The right hand side is a perfect square, so we can write:
2 y x^2  2 x + y^2 + 8 =
2 y (x^2  1/y x + y/2 + 4/y) =
2 y [x  1/(2y)]^2
So, we have:
(x^2 + y)^2 = 2 y [x  1/(2y)]^2
Take square roots of bth sides:
x^2 + y = ± sqrt(2y) (x + 1/(2y))
And this is just an ordinary quadratic equation that you can solve using the quadratic formula! The ± sign means that there are two quadric equation, each has (in general) two solutions, so in general you get 4 solutions.
The question then remains of how to solve that third degree equation for y:
y^3 + 8 y  1/2 = 0
You can do this by writing:
y^3 = 8 y + 1/2
You can compare this to what you get when you expand out (a+b)^3:
(a+b)^3 = a^3 + b^3 + 3 a^2b + 3 a b^2
The right hand side can be written as:
3 a b (a + b) + a^3 + b^3.
So, we have the identity:
(a+b)^3 = 3 a b (a + b) + a^3 + b^3.
This is obviously always valid, no matter what a and b are. But this looks similar to the equation we want to solve:
y^3 = 8 y + 1/2
with y playing the role of a + b. Suppose then we somehow manage to find an a and b such that
3 a b = 8
and
a^3 + b^3 = 1/2
Then because
(a + b)^3 = 3 a b (a + b) + a^3 + b^3
is always valid, it is also valid for this particular case. But then:
(a + b)^3 = 8 (a + b) + 1/2
which means that y = a + b then satisfies the equation we want to solve:
y^3 = 8 y + 1/2
Finding a and b, involves nothing more than solving a quadratic equation.
Taking te third power of the equation
3 a b = 8
gives:
27 a^3 b^3 =  512
And we also have the equation:
a^3 + b^3 = 1/2
Put:
A = a^3
B = b^3
Then you have the equations:
A B = 512/27
A + B = 1/2
If you use the last one to express B in terms of A and plug that into the first one, you obtain a quadratic equation for A.
Then from A and B you compute a and b and then y is given as a + b.
So, you see that solving third and fourth degree equations is easy, you only need to use same kind of algebra that goes into solving quadratic equations.