Solution - Linear equations with one unknown
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "c5" was replaced by "c^5".
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
2*c^2-9*c^5-(2*c)=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((2 • (c2)) - 32c5) - 2c = 0Step 2 :
Equation at the end of step 2 :
(2c2 - 32c5) - 2c = 0
Step 3 :
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
-9c5 + 2c2 - 2c = -c • (9c4 - 2c + 2)
Polynomial Roots Calculator :
4.2 Find roots (zeroes) of : F(c) = 9c4 - 2c + 2
Polynomial Roots Calculator is a set of methods aimed at finding values of c for which F(c)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers c 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 9 and the Trailing Constant is 2.
The factor(s) are:
of the Leading Coefficient : 1,3 ,9
of the Trailing Constant : 1 ,2
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 13.00 | ||||||
| -1 | 3 | -0.33 | 2.78 | ||||||
| -1 | 9 | -0.11 | 2.22 | ||||||
| -2 | 1 | -2.00 | 150.00 | ||||||
| -2 | 3 | -0.67 | 5.11 | ||||||
| -2 | 9 | -0.22 | 2.47 | ||||||
| 1 | 1 | 1.00 | 9.00 | ||||||
| 1 | 3 | 0.33 | 1.44 | ||||||
| 1 | 9 | 0.11 | 1.78 | ||||||
| 2 | 1 | 2.00 | 142.00 | ||||||
| 2 | 3 | 0.67 | 2.44 | ||||||
| 2 | 9 | 0.22 | 1.58 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 4 :
-c • (9c4 - 2c + 2) = 0
Step 5 :
Theory - Roots of a product :
5.1 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 :
5.2 Solve : -c = 0
Multiply both sides of the equation by (-1) : c = 0
Quartic Equations :
5.3 Solve 9c4-2c+2 = 0
In search of an interavl at which the above polynomial changes sign, from negative to positive or the other wayaround.
Method of search: Calculate polynomial values for all integer points between c=-20 and c=+20
No interval at which a change of sign occures has been found. Consequently, Bisection Approximation can not be used. As this is a polynomial of an even degree it may not even have any real (as opposed to imaginary) roots
One solution was found :
c = 0How did we do?
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