# Math

When you use the quadratic formula (x= -b+/=......) can you use that on equations like X^4+2X-8?

If not, how would you algebraically find the roots of: X^4+2X-8?

Thanks!

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1. 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

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2. 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.

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