Solution - Adding, subtracting and finding the least common multiple
Other Ways to Solve
Adding, subtracting and finding the least common multipleStep by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "0.00523224" was replaced by "(00523224/100000000)". 3 more similar replacement(s)
Step by step solution :
Step 1 :
65403
Simplify ————————
12500000
Equation at the end of step 1 :
34545 1352 65403
(((x3)+(—————•(x2)))+(—————•x))-———————— = 0
10000 10000 12500000
Step 2 :
169
Simplify ————
1250
Equation at the end of step 2 :
34545 169 65403 (((x3)+(—————•(x2)))+(————•x))-———————— = 0 10000 1250 12500000Step 3 :
6909 Simplify ———— 2000
Equation at the end of step 3 :
6909 169x 65403
(((x3) + (———— • x2)) + ————) - ———————— = 0
2000 1250 12500000
Step 4 :
Equation at the end of step 4 :
6909x2 169x 65403 (((x3) + ——————) + ————) - ———————— = 0 2000 1250 12500000Step 5 :
Rewriting the whole as an Equivalent Fraction :
5.1 Adding a fraction to a whole
Rewrite the whole as a fraction using 2000 as the denominator :
x3 x3 • 2000
x3 = —— = —————————
1 2000
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 :
5.2 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:
x3 • 2000 + 6909x2 2000x3 + 6909x2
—————————————————— = ———————————————
2000 2000
Equation at the end of step 5 :
(2000x3 + 6909x2) 169x 65403
(————————————————— + ————) - ———————— = 0
2000 1250 12500000
Step 6 :
Step 7 :
Pulling out like terms :
7.1 Pull out like factors :
2000x3 + 6909x2 = x2 • (2000x + 6909)
Calculating the Least Common Multiple :
7.2 Find the Least Common Multiple
The left denominator is : 2000
The right denominator is : 1250
| Prime Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
|---|---|---|---|
| 2 | 4 | 1 | 4 |
| 5 | 3 | 4 | 4 |
| Product of all Prime Factors | 2000 | 1250 | 10000 |
Least Common Multiple:
10000
Calculating Multipliers :
7.3 Calculate multipliers for the two fractions
Denote the Least Common Multiple by L.C.M
Denote the Left Multiplier by Left_M
Denote the Right Multiplier by Right_M
Denote the Left Deniminator by L_Deno
Denote the Right Multiplier by R_Deno
Left_M = L.C.M / L_Deno = 5
Right_M = L.C.M / R_Deno = 8
Making Equivalent Fractions :
7.4 Rewrite the two fractions into equivalent fractions
Two fractions are called equivalent if they have the same numeric value.
For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.
To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.
L. Mult. • L. Num. x2 • (2000x+6909) • 5 —————————————————— = ————————————————————— L.C.M 10000 R. Mult. • R. Num. 169x • 8 —————————————————— = ———————— L.C.M 10000
Adding fractions that have a common denominator :
7.5 Adding up the two equivalent fractions
x2 • (2000x+6909) • 5 + 169x • 8 10000x3 + 34545x2 + 1352x
———————————————————————————————— = —————————————————————————
10000 10000
Equation at the end of step 7 :
(10000x3 + 34545x2 + 1352x) 65403
——————————————————————————— - ———————— = 0
10000 12500000
Step 8 :
Step 9 :
Pulling out like terms :
9.1 Pull out like factors :
10000x3 + 34545x2 + 1352x = x • (10000x2 + 34545x + 1352)
Trying to factor by splitting the middle term
9.2 Factoring 10000x2 + 34545x + 1352
The first term is, 10000x2 its coefficient is 10000 .
The middle term is, +34545x its coefficient is 34545 .
The last term, "the constant", is +1352
Step-1 : Multiply the coefficient of the first term by the constant 10000 • 1352 = 13520000
Step-2 : Find two factors of 13520000 whose sum equals the coefficient of the middle term, which is 34545 .
Numbers too big. Method shall not be applied
Calculating the Least Common Multiple :
9.3 Find the Least Common Multiple
The left denominator is : 10000
The right denominator is : 12500000
| Prime Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
|---|---|---|---|
| 2 | 4 | 5 | 5 |
| 5 | 4 | 8 | 8 |
| Product of all Prime Factors | 10000 | 12500000 | 12500000 |
Least Common Multiple:
12500000
Calculating Multipliers :
9.4 Calculate multipliers for the two fractions
Denote the Least Common Multiple by L.C.M
Denote the Left Multiplier by Left_M
Denote the Right Multiplier by Right_M
Denote the Left Deniminator by L_Deno
Denote the Right Multiplier by R_Deno
Left_M = L.C.M / L_Deno = 1250
Right_M = L.C.M / R_Deno = 1
Making Equivalent Fractions :
9.5 Rewrite the two fractions into equivalent fractions
L. Mult. • L. Num. x • (10000x2+34545x+1352) • 1250 —————————————————— = ———————————————————————————————— L.C.M 12500000 R. Mult. • R. Num. 65403 —————————————————— = ———————— L.C.M 12500000
Adding fractions that have a common denominator :
9.6 Adding up the two equivalent fractions
x • (10000x2+34545x+1352) • 1250 - (65403) 12500000x3 + 43181250x2 + 1690000x - 65403
—————————————————————————————————————————— = ——————————————————————————————————————————
12500000 12500000
Checking for a perfect cube :
9.7 12500000x3 + 43181250x2 + 1690000x - 65403 is not a perfect cube
Trying to factor by pulling out :
9.8 Factoring: 12500000x3 + 43181250x2 + 1690000x - 65403
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 43181250x2 - 65403
Group 2: 12500000x3 + 1690000x
Pull out from each group separately :
Group 1: (14393750x2 - 21801) • (3)
Group 2: (1250x2 + 169) • (10000x)
Bad news !! Factoring by pulling out fails :
The groups have no common factor and can not be added up to form a multiplication.
Polynomial Roots Calculator :
9.9 Find roots (zeroes) of : F(x) = 12500000x3 + 43181250x2 + 1690000x - 65403
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 12500000 and the Trailing Constant is -65403.
The factor(s) are:
of the Leading Coefficient : 1,2 ,4 ,5 ,8 ,10 ,16 ,20 ,25 ,32 , etc
of the Trailing Constant : 1 ,3 ,9 ,13 ,39 ,43 ,117 ,129 ,169 ,387 , etc
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 28925847.00 | ||||||
| -1 | 2 | -0.50 | 8322409.50 | ||||||
| -1 | 4 | -0.25 | 2015612.62 | ||||||
| -1 | 5 | -0.20 | 1223847.00 | ||||||
| -1 | 8 | -0.12 | 373639.97 |
Note - For tidiness, printing of 195 checks which found no root was suppressed
Polynomial Roots Calculator found no rational roots
Equation at the end of step 9 :
12500000x3 + 43181250x2 + 1690000x - 65403
—————————————————————————————————————————— = 0
12500000
Step 10 :
When a fraction equals zero :
10.1 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:
12500000x3+43181250x2+1690000x-65403
———————————————————————————————————— • 12500000 = 0 • 12500000
12500000
Now, on the left hand side, the 12500000 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
12500000x3+43181250x2+1690000x-65403 = 0
Cubic Equations :
10.2 Solve 12500000x3+43181250x2+1690000x-65403 = 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) = 12500000x3 + 43181250x2 + 1690000x - 65403
At x= -4.00 F(x) is equal to -115925403.00
At x= -3.00 F(x) is equal to 45995847.00
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) -4.000000000 -115925403.00 -3.000000000 45995847.00 0.000000000 -65403.00 -3.000000000 45995847.00 0.000000000 -65403.00 -1.500000000 52369909.50 0.000000000 -65403.00 -0.750000000 17683112.62 0.000000000 -65403.00 -0.375000000 4714030.59 0.000000000 -65403.00 -0.187500000 1053415.36 0.000000000 -65403.00 -0.093750000 145382.52 -0.046875000 -51028.53 -0.093750000 145382.52 -0.046875000 -51028.53 -0.070312500 24905.22 -0.058593750 -18689.95 -0.070312500 24905.22 -0.058593750 -18689.95 -0.064453125 1708.103 -0.061523438 -8841.750 -0.064453125 1708.103 -0.062988281 -3654.412 -0.064453125 1708.103 -0.063720703 -995.036483868 -0.064453125 1708.103 -0.063720703 -995.036483868 -0.064086914 351.064696193 -0.063903809 -323.353312732 -0.064086914 351.064696193 -0.063903809 -323.353312732 -0.063995361 13.513865784 -0.063949585 -155.005183558 -0.063995361 13.513865784 -0.063972473 -70.767023458 -0.063995361 13.513865784 -0.063983917 -28.631919924 -0.063995361 13.513865784 -0.063989639 -7.560362335 -0.063995361 13.513865784 -0.063989639 -7.560362335 -0.063992500 2.976417909 -0.063991070 -2.292055667 -0.063992500 2.976417909 -0.063991070 -2.292055667 -0.063991785 0.342160258 -0.063991427 -0.974952920 -0.063991785 0.342160258 -0.063991606 -0.316397635 -0.063991785 0.342160258 -0.063991606 -0.316397635 -0.063991696 0.012880986 -0.063991651 -0.151758406 -0.063991696 0.012880986 -0.063991673 -0.069438731 -0.063991696 0.012880986 -0.063991684 -0.028278878 -0.063991696 0.012880986 -0.063991690 -0.007698947 -0.063991696 0.012880986 -0.063991690 -0.007698947 -0.063991693 0.002591019 -0.063991691 -0.002553964 -0.063991693 0.002591019 -0.063991691 -0.002553964 -0.063991692 0.000018527 -0.063991692 -0.001267719 -0.063991692 0.000018527 -0.063991692 -0.000624596 -0.063991692 0.000018527 -0.063991692 -0.000303034 -0.063991692 0.000018527 -0.063991692 -0.000142254 -0.063991692 0.000018527 -0.063991692 -0.000061863 -0.063991692 0.000018527 -0.063991692 -0.000021668 -0.063991692 0.000018527 -0.063991692 -0.000001570 -0.063991692 0.000018527 -0.063991692 -0.000001570 -0.063991692 0.000008478
Next Middle will get us close enough to zero:
F( -0.063991692 ) is 0.000000942
The desired approximation of the solution is:
x ≓ -0.063991692
Note, ≓ is the approximation symbol
One solution was found :
x ≓ -0.063991692How did we do?
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