Solution - Approximation

Rearrange:

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

                     x^3-(x^2-7)=0 

Step by step solution :

Step  1  :

Polynomial Roots Calculator :

 1.1    Find roots (zeroes) of :       F(x) = x³-x²+7
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  7.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,7

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  1  :

  x3 - x2 + 7  = 0 

Step  2  :

Cubic Equations :

 2.1     Solve   x³-x²+7 = 0

Future releases of Tiger-Algebra will solve equations of the third degree directly.

Meanwhile we will use the Bisection method to approximate one real solution.

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) = x³ - x² + 7

At   x=   -2.00   F(x)  is equal to  -5.00 
At   x=   -1.00   F(x)  is equal to  5.00 

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   -5.000000000 -1.000000000    5.000000000
-2.000000000   -5.000000000  0.000000000    7.000000000
-2.000000000   -5.000000000 -1.000000000    5.000000000
-2.000000000   -5.000000000 -1.500000000    1.375000000
-1.750000000   -1.421875000 -1.500000000    1.375000000
-1.750000000   -1.421875000 -1.625000000    0.068359375
-1.687500000   -0.653076172 -1.625000000    0.068359375
-1.656250000   -0.286529541 -1.625000000    0.068359375
-1.640625000   -0.107639313 -1.625000000    0.068359375
-1.632812500   -0.019279957 -1.625000000    0.068359375
-1.632812500   -0.019279957 -1.628906250    0.024629533
-1.632812500   -0.019279957 -1.630859375    0.002697267
-1.631835938   -0.008285723 -1.630859375    0.002697267
-1.631347656   -0.002792823 -1.630859375    0.002697267
-1.631103516   -0.000047427 -1.630859375    0.002697267
-1.631103516   -0.000047427 -1.630981445    0.001325008
-1.631103516   -0.000047427 -1.631042480    0.000638812
-1.631103516   -0.000047427 -1.631072998    0.000295698
-1.631103516   -0.000047427 -1.631088257    0.000124137
-1.631103516   -0.000047427 -1.631095886    0.000038356
-1.631099701   -0.000004536 -1.631095886    0.000038356
-1.631099701   -0.000004536 -1.631097794    0.000016910

     Next Middle will get us close enough to zero:

     F( -1.631099224 ) is   0.000000826  

     The desired approximation of the solution is:

       x ≓ -1.631099224

     Note, ≓ is the approximation symbol

One solution was found :