Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "1.4" was replaced by "(14/10)". 2 more similar replacement(s)

Rearrange:

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

                (38/10)*n-13-((14/10)*n^5)=0 

Step by step solution :

Step  1  :

            7
 Simplify   —
            5

Equation at the end of step  1  :

    38                 7
  ((—— • n) -  13) -  (— • n5)  = 0 
    10                 5

Step  2  :

Equation at the end of step  2  :

    38                7n5
  ((—— • n) -  13) -  ———  = 0 
    10                 5 

Step  3  :

            19
 Simplify   ——
            5 

Equation at the end of step  3  :

    19                7n5
  ((—— • n) -  13) -  ———  = 0 
    5                  5 

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  5  as the denominator :

          13     13 • 5
    13 =  ——  =  ——————
          1        5   

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 4.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 19n - (13 • 5)     19n - 65
 ——————————————  =  ————————
       5               5    

Equation at the end of step  4  :

  (19n - 65)    7n5
  —————————— -  ———  = 0 
      5          5 

Step  5  :

Adding fractions which have a common denominator :

 5.1       Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 (19n-65) - (7n5)     -7n5 + 19n - 65
 ————————————————  =  ———————————————
        5                    5       

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   -7n⁵ + 19n - 65  =   -1 • (7n⁵ - 19n + 65) 

Polynomial Roots Calculator :

 6.2    Find roots (zeroes) of :       F(n) = 7n⁵ - 19n + 65
Polynomial Roots Calculator is a set of methods aimed at finding values of  n  for which   F(n)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  n  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  7  and the Trailing Constant is  65.

 The factor(s) are:

of the Leading Coefficient :  1,7
 of the Trailing Constant :  1 ,5 ,13 ,65

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  6  :

  -7n5 + 19n - 65
  ———————————————  = 0 
         5       

Step  7  :

When a fraction equals zero :

 7.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  -7n5+19n-65
  ——————————— • 5 = 0 • 5
       5     

Now, on the left hand side, the  5  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -7n⁵+19n-65  = 0

Equations of order 5 or higher :

 7.2     Solve   -7n⁵+19n-65 = 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(n) = -7n⁵ + 19n - 65

At   n=   -1.00   F(n)  is equal to  -77.00 
At   n=   -2.00   F(n)  is equal to  121.00 

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

-1.000000000  -77.000000000 -2.000000000  121.000000000
 0.000000000  -65.000000000 -2.000000000  121.000000000
-1.000000000  -77.000000000 -2.000000000  121.000000000
-1.500000000  -40.343750000 -2.000000000  121.000000000
-1.500000000  -40.343750000 -1.750000000   16.641601562
-1.625000000  -16.558258057 -1.750000000   16.641601562
-1.687500000   -1.273211479 -1.750000000   16.641601562
-1.687500000   -1.273211479 -1.718750000    7.337052375
-1.687500000   -1.273211479 -1.703125000    2.947490613
-1.687500000   -1.273211479 -1.695312500    0.816321902
-1.691406250   -0.233613271 -1.695312500    0.816321902
-1.691406250   -0.233613271 -1.693359375    0.290057716
-1.691406250   -0.233613271 -1.692382812    0.027898633
-1.691894531   -0.102938146 -1.692382812    0.027898633
-1.692138672   -0.037539972 -1.692382812    0.027898633
-1.692260742   -0.004825724 -1.692382812    0.027898633
-1.692260742   -0.004825724 -1.692321777    0.011535191
-1.692260742   -0.004825724 -1.692291260    0.003354417
-1.692276001   -0.000735732 -1.692291260    0.003354417
-1.692276001   -0.000735732 -1.692283630    0.001309323
-1.692276001   -0.000735732 -1.692279816    0.000286790
-1.692277908   -0.000224472 -1.692279816    0.000286790
-1.692277908   -0.000224472 -1.692278862    0.000031159
-1.692278385   -0.000096657 -1.692278862    0.000031159

     Next Middle will get us close enough to zero:

     F( -1.692278743 ) is  -0.000000795  

     The desired approximation of the solution is:

       n ≓ -1.692278743

     Note, ≓ is the approximation symbol

One solution was found :