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!
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
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.
No, the quadratic formula cannot be directly used to find the roots of equations like X^4 + 2X - 8. The quadratic formula is specifically designed to find the roots of quadratic equations in the form ax^2 + bx + c = 0, where a, b, and c are coefficients.
To algebraically find the roots of the equation X^4 + 2X - 8, we can use other methods like factoring, the rational root theorem, or numerical methods like Newton's method.
1. Factoring: Unfortunately, in this case, it is not possible to easily factor the equation X^4 + 2X - 8 into a product of linear factors, so this method doesn't work.
2. Rational Root Theorem: The rational root theorem states that if a rational number p/q is a root of a polynomial equation with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.
In this case, the leading coefficient is 1, and the constant term is -8. The factors of -8 are ±1, ±2, ±4, and ±8, and the factors of 1 are ±1. By testing these possible rational roots in the equation, we can check if any of them satisfy the equation.
Unfortunately, it can be quite time-consuming and may not yield any rational roots for higher-degree polynomials.
3. Numerical Methods: For higher-degree polynomial equations like this one, numerical methods are often used to approximate the roots. One commonly used numerical method is Newton's method, which involves applying iterative calculations to approximate the roots. However, this method typically requires initial guesses for the roots and involves more advanced mathematical techniques.
In some cases, advanced mathematical methods such as the use of special functions or numerical algorithms may be necessary to find the roots of higher-degree polynomials. However, there is no general formula like the quadratic formula for equations of higher degrees.