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Solution - Approximation

x ≓ - 1.267168283

x≓-1.267168283

x = 0

x=0

Step by Step Solution

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

x*(x-2)-((x^3)^2)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

  x • (x - 2) -  x⁶  = 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

-x⁶ + x² - 2x = -x • (x⁵ - x + 2)

Polynomial Roots Calculator :

3.2    Find roots (zeroes) of :       F(x) = x⁵ - x + 2
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 1 and the Trailing Constant is 2.

The factor(s) are:

of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2

Let us test ....

P	Q	P/Q	F(P/Q)
-1	1	-1.00	2.00
-2	1	-2.00	-28.00
1	1	1.00	2.00
2	1	2.00	32.00

Polynomial Roots Calculator found no rational roots

Equation at the end of step 3 :

  -x • (x⁵ - x + 2)  = 0

Step 4 :

Theory - Roots of a product :

4.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

4.2 Solve : -x = 0

Multiply both sides of the equation by (-1) : x = 0

Equations of order 5 or higher :

4.3 Solve x⁵-x+2 = 0

Points regarding equations of degree five or higher.

(1) There is no general method (Formula) for solving polynomial equations of degree five or higher.

(2) By the Fundamental theorem of Algebra, if we allow complex numbers, an equation of degree n will have exactly n solutions
(This is if we count double solutions as 2 , triple solutions as 3 and so on

) (3) By the Abel-Ruffini theorem, the solutions can not always be presented in the conventional way using only a finite amount of additions, subtractions, multiplications, divisions or root extractions

(4) If F(x) is a polynomial of odd degree with real coefficients, then the equation F(X)=0 has at least one real solution.

(5) Using methods such as the Bisection Method, real solutions can be approximated to any desired degree of accuracy.

Approximating a root using the Bisection Method :

We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).

The function is F(x) = x⁵ - x + 2

At x= -2.00 F(x) is equal to -28.00
At x= -1.00 F(x) is equal to 2.00

Intuitively we feel, and justly so, that since F(x) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(x) is zero

Procedure :
(1) Find a point "Left" where F(Left) < 0

(2) Find a point 'Right' where F(Right) > 0

(3) Compute 'Middle' the middle point of the interval [Left,Right]

(4) Calculate Value = F(Middle)

(5) If Value is close enough to zero goto Step (7)

Else :
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle

(6) Loop back to Step (3)

(7) Done!! The approximation found is Middle

Follow Middle movements to understand how it works :

    Left       Value(Left)     Right       Value(Right)

-2.000000000  -28.000000000 -1.000000000    2.000000000
-2.000000000  -28.000000000  0.000000000    2.000000000
-2.000000000  -28.000000000 -1.000000000    2.000000000
-1.500000000   -4.093750000 -1.000000000    2.000000000
-1.500000000   -4.093750000 -1.250000000    0.198242188
-1.375000000   -1.539886475 -1.250000000    0.198242188
-1.312500000   -0.582402229 -1.250000000    0.198242188
-1.281250000   -0.171533853 -1.250000000    0.198242188
-1.281250000   -0.171533853 -1.265625000    0.018303975
-1.273437500   -0.075354502 -1.265625000    0.018303975
-1.269531250   -0.028213050 -1.265625000    0.018303975
-1.267578125   -0.004876844 -1.265625000    0.018303975
-1.267578125   -0.004876844 -1.266601562    0.006732944
-1.267578125   -0.004876844 -1.267089844    0.000932900
-1.267333984   -0.001970759 -1.267089844    0.000932900
-1.267211914   -0.000518626 -1.267089844    0.000932900
-1.267211914   -0.000518626 -1.267150879    0.000207213
-1.267181396   -0.000155688 -1.267150879    0.000207213
-1.267181396   -0.000155688 -1.267166138    0.000025767
-1.267173767   -0.000064959 -1.267166138    0.000025767
-1.267169952   -0.000019596 -1.267166138    0.000025767
-1.267169952   -0.000019596 -1.267168045    0.000003086
-1.267168999   -0.000008255 -1.267168045    0.000003086

     Next Middle will get us close enough to zero:

     F( -1.267168283 ) is 0.000000251

     The desired approximation of the solution is:

       x ≓ -1.267168283

     Note, ≓ is the approximation symbol

Two solutions were found :

x ≓ -1.267168283
x = 0

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