Lesson Solving algebraic equations of high degree

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This Lesson (Solving algebraic equations of high degree) was created by by ikleyn(4) About Me : View Source, Show
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Some algebraic equations of high degree can be solved by reduction to the quadratic equation.
Below are examples of three forms of such equations.
Note that lessons explaining how to solve quadratic equations are PROOF of quadratic formula..., Introduction into quadratic equations and Solving quadratic equations without quadratic formula of the module Quadratic equations.

Example 1.
Solve an equation x%5E4%2B5x%5E3%2B6x%5E2=0.

Polynomial x%5E4%2B5x%5E3%2B6x%5E2=0 can be factored into the product of x%5E2 and %28x%5E2%2B5x%2B6%29.
Solve an equation x%5E2=0 first. It has two equal roots:x%5B1%5D=x%5B2%5D=0.
Then solve an equation x%5E2%2B5x%2B6=0. It has roots x%5B3%5D=-2, x%5B4%5D=-3.
Thus, roots of the original equation are x%5B1%5D=x%5B2%5D=0, x%5B3%5D=-2, x%5B4%5D=-3.

Example 2.
Solve an equation x%5E3=8.

Let's rewrite this equation as x%5E3-8=0.
One root is x%5B1%5D=2. Factor left side of x%5E3-8 into the product of
x-2 and the polynomial of degree 2 x%5E2%2B2x%2B4, so
x%5E3-8=%28x-2%29%2A%28x%5E2%2B2x%2B4%29.
Solve an equation x%5E2%2B2x%2B4=0 using quadratic formula. The two roots are complex numbers
x%5B2%5D=-1%2Bsqrt%28-3%29 and x%5B3%5D=-1-sqrt%28-3%29.
Thus, roots of the original equation are x%5B1%5D=2, x%5B2%5D=-1%2Bsqrt%28-3%29 and x%5B3%5D=-1-sqrt%28-3%29.

Example 3.
Solve an equation x%5E4-13x%5E2%2B36=0.

Let's introduce new variable z=x%5E2. Then the equation takes the form z%5E2-13z%2B36=0.
This is the quadratic equation. It has two roots z%5B1%5D=9, z%5B2%5D=4.
Now solve an equation x%5E3=9. The roots of this equation are x%5B1%5D=3 and x%5B1%5D=-3.
Then solve an equation x%5E3=4. The roots of this equation are x%5B1%5D=2 and x%5B1%5D=-2.
So, the roots of the original equation are x%5B1%5D=3, x%5B1%5D=-3, x%5B3%5D=2 and x%5B4%5D=-2.

Equations of this form are called bi-quadratic equations.

Example 4.
Solve an equation x%5E6-16x%5E3%2B64=0.

Let's introduce new variable z=x%5E3. Then the equation takes the form z%5E2-16z%2B64=0.
This is the quadratic equation. It has two equal roots z%5B1%5D=z%5B2%5D=8.
Now solve an equation x%5E3=8. It has three roots
x%5B1%5D=2, x%5B2%5D-1%2Bsqrt%28-3%29 and x%5B3%5D=-1-sqrt%28-3%29 (see Example 2).
They are solutions of the original equation.
Other three roots of the original equation are equal to these three (because z%5B1%5D=z%5B2%5D).

Method of Example 1 can be extended to solve equations of the form
x%5En%2A%28ax%5E2%2Bbx%2Bc%29=0.

Method of Example 2 can be extended to solve equations of the form
%28ax%2Bb%29%2A%28cx%5E2%2Bdx%2Be%29=0.

Method of Example 3 can be extended to solve equations of the form
ax%5E%282n%29%2Bbx%5En%2Bc=0.

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