Solution - Approximation
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 :
25*x^2-500/x-(25*x)=0
Step by step solution :
Step 1 :
500
Simplify ———
x
Equation at the end of step 1 :
500 ((25 • (x2)) - ———) - 25x = 0 xStep 2 :
Equation at the end of step 2 :
500
(52x2 - ———) - 25x = 0
x
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using x as the denominator :
52x2 52x2 • x
52x2 = ———— = ————————
1 x
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 :
3.2 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:
52x2 • x - (500) 25x3 - 500
———————————————— = ——————————
x x
Equation at the end of step 3 :
(25x3 - 500)
———————————— - 25x = 0
x
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using x as the denominator :
25x 25x • x
25x = ——— = ———————
1 x
Step 5 :
Pulling out like terms :
5.1 Pull out like factors :
25x3 - 500 = 25 • (x3 - 20)
Trying to factor as a Difference of Cubes:
5.2 Factoring: x3 - 20
Theory : A difference of two perfect cubes, a3 - b3 can be factored into
(a-b) • (a2 +ab +b2)
Proof : (a-b)•(a2+ab+b2) =
a3+a2b+ab2-ba2-b2a-b3 =
a3+(a2b-ba2)+(ab2-b2a)-b3 =
a3+0+0-b3 =
a3-b3
Check : 20 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Polynomial Roots Calculator :
5.3 Find roots (zeroes) of : F(x) = x3 - 20
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 -20.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -21.00 | ||||||
| -2 | 1 | -2.00 | -28.00 | ||||||
| -4 | 1 | -4.00 | -84.00 | ||||||
| -5 | 1 | -5.00 | -145.00 | ||||||
| -10 | 1 | -10.00 | -1020.00 | ||||||
| -20 | 1 | -20.00 | -8020.00 | ||||||
| 1 | 1 | 1.00 | -19.00 | ||||||
| 2 | 1 | 2.00 | -12.00 | ||||||
| 4 | 1 | 4.00 | 44.00 | ||||||
| 5 | 1 | 5.00 | 105.00 | ||||||
| 10 | 1 | 10.00 | 980.00 | ||||||
| 20 | 1 | 20.00 | 7980.00 |
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
5.4 Adding up the two equivalent fractions
25 • (x3-20) - (25x • x) 25x3 - 25x2 - 500
———————————————————————— = —————————————————
x x
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
25x3 - 25x2 - 500 = 25 • (x3 - x2 - 20)
Polynomial Roots Calculator :
6.2 Find roots (zeroes) of : F(x) = x3 - x2 - 20
See theory in step 5.3
In this case, the Leading Coefficient is 1 and the Trailing Constant is -20.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -22.00 | ||||||
| -2 | 1 | -2.00 | -32.00 | ||||||
| -4 | 1 | -4.00 | -100.00 | ||||||
| -5 | 1 | -5.00 | -170.00 | ||||||
| -10 | 1 | -10.00 | -1120.00 | ||||||
| -20 | 1 | -20.00 | -8420.00 | ||||||
| 1 | 1 | 1.00 | -20.00 | ||||||
| 2 | 1 | 2.00 | -16.00 | ||||||
| 4 | 1 | 4.00 | 28.00 | ||||||
| 5 | 1 | 5.00 | 80.00 | ||||||
| 10 | 1 | 10.00 | 880.00 | ||||||
| 20 | 1 | 20.00 | 7580.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 6 :
25 • (x3 - x2 - 20)
——————————————————— = 0
x
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:
25•(x3-x2-20)
————————————— • x = 0 • x
x
Now, on the left hand side, the x cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
25 • (x3-x2-20) = 0
Equations which are never true :
7.2 Solve : 25 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Cubic Equations :
7.3 Solve x3-x2-20 = 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 - 20
At x= 3.00 F(x) is equal to -2.00
At x= 4.00 F(x) is equal to 28.00
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) 3.000000000 -2.000000000 4.000000000 28.000000000 0.000000000 -20.000000000 4.000000000 28.000000000 2.000000000 -16.000000000 4.000000000 28.000000000 3.000000000 -2.000000000 4.000000000 28.000000000 3.000000000 -2.000000000 3.500000000 10.625000000 3.000000000 -2.000000000 3.250000000 3.765625000 3.000000000 -2.000000000 3.125000000 0.751953125 3.062500000 -0.656005859 3.125000000 0.751953125 3.062500000 -0.656005859 3.093750000 0.039886475 3.078125000 -0.310070038 3.093750000 0.039886475 3.085937500 -0.135595798 3.093750000 0.039886475 3.089843750 -0.047980845 3.093750000 0.039886475 3.091796875 -0.004078753 3.093750000 0.039886475 3.091796875 -0.004078753 3.092773438 0.017895966 3.091796875 -0.004078753 3.092285156 0.006906633 3.091796875 -0.004078753 3.092041016 0.001413447 3.091918945 -0.001332777 3.092041016 0.001413447 3.091918945 -0.001332777 3.091979980 0.000040304 3.091949463 -0.000646244 3.091979980 0.000040304 3.091964722 -0.000302972 3.091979980 0.000040304 3.091972351 -0.000131334 3.091979980 0.000040304 3.091976166 -0.000045515 3.091979980 0.000040304 3.091978073 -0.000002606 3.091979980 0.000040304 3.091978073 -0.000002606 3.091979027 0.000018849 3.091978073 -0.000002606 3.091978550 0.000008122
Next Middle will get us close enough to zero:
F( 3.091978192 ) is 0.000000076
The desired approximation of the solution is:
x ≓ 3.091978192
Note, ≓ is the approximation symbol
One solution was found :
x ≓ 3.091978192How did we do?
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