The algorithm was presented without proof to avoid others claiming the idea as their own. The method is clearly the same as the formula method of solution. He showed the solution of particular equations, for example: x2 - 10x = -9. He called the numbers by the names:
1 is "the square"
-10 is "the unknown"
-9 is "the absolute number"
Brahmagupta's work yields only one solution for each quadratic equation. However, by the 12th century, scholars were identifying both solutions. Problems were now being written to teach mental skills rather than to solve problems from astronomy.
x - 2.3) = x2- 3.8x + 3.45 the quadratic function can be form
This is a quadratic function.
When x is 3 this quadratic function is zero.
x2 - 4x + 3
x = 3
32 - 4 × 3 + 3
9 - 12 + 3
The Factor Theorem states that if a function of xis zero when x = 3, then (x - 3) is a factor of the function.
f(3) = 0 (x - 3) is a factor of f(x) = x2 - 4x + 3
The other factor can then be found to give the completely factorised form of the quadratic.
By multiplication it is found that the other factor is (x -1) and f(x) = (x - 3)(x -1)
The Factor Theorem states that if f(a) = 0 then (x - a) is a factor of the function f(x).
(x - a) is a factor of f(x) because f(a) = 0
The process can be used in reverse to find a quadratic function with particular properties. Find the function with factors (x - 3) and (x + 5)
The function with factors (x -3) and (x + 5) is
(x - 3)(x + 5)
x2 + 2x - 15
Sometimes functions with higher powers of x can be factorised by this method .
The process of obtaining the factors becomes longer and sometimes involves long-division.
f(x) = x3 - 12x2 + 47x - 60 f(5) = 0 (x - 5) is a factor.
Long division takes practise. Here is an example:
Divide: f(x) = x3 - 12x2 + 47x - 60 by the factor (x -5) to find the other factors
x cubed divided by x gives x squared multiply the factor (x -5) by x2 to give x3 - 5x2 subtract this from the first two terms and then divide the answer by x This gives -7x and after multiplication 7x2 + 47x Subtract to leave 12x which when divided by x gives 12
Long division is not essential. Other factors can be found using the Factor Theorem again. Factorising continues with the quadratic factor.
x1 = 0.5 x2 = 1 x3 = -1.148698 x4 = -1.268010 x5 = -1.346816 I can immediately see that there is no convergence in this rearrangement towards the particular root I am looking for, that I found in the previous rearrangement (Rearrangement 2).
is 1. I will now draw a graph that will also show the maximum amount for x and to verify that x is 1 and that the greatest volume possible is 16. The graph shows also shows that the maximum volume is about 16 and x is 1.
x y = x5-3x+1 0.33 0.013914 0.331 0.010973 0.332 0.008034 0.333 0.005095 0.334 0.002157 0.335 -0.00078 x y = x5-3x+1 0.334 0.002157 0.3341 0.001863 0.3342 0.001569 0.3343 0.001275 0.3344 0.000981 0.3345 0.000688 0.3346 0.000394 0.3347 0.0001 0.3348 -0.00019 I am continuing this process of decimal search until I find the
After finding the transpose, we need to find the co-factors of the matrix. This is achieved by taking the transpose, getting each number of the matrix - eliminating the corresponding rows and columns and then cross multiplying and the subtracting what remains. [(15X2-6X4) (2X5-10X4) (5X6-10X15) ] [(2X10-6X8) (2X20-10X8) (6X20-10X10)] [(4X10-15X8)
Formula For Connect 4 The formula for any box with a width of 4 is C=7H-18 The formula for any box with a width of 5 is C=11H-27 The formula for any box with a width of 6 is C=15H-36 There is a visible pattern the first number always goes
Assuming it does we will be able to approximate the root for another value of K. It will only be a approximation as the value of our roots are approximations. To prove that there is a common factor, C we do the following: (assume k=(k+1)
If I only had a scientific calculator, Newton Raphson will be the fastest way to find the roots. The reason is that it only involves a few calculations and it is easy to calculate them. On the other hand, change of signs method involves a number of tables and it makes it inconvenience for us to find the roots.
In order to start the iteration, you need to start with a starting value, or seed, that takes the form of z0. Thus, iterating this starting value with the function yields: z1 = (z0)2 + z0 z2 = (z1)2 + z0 z3 = (z2)2 + z0 z4 = (z3)2 + z0 z5 = (z4)2 + z0 and so on.
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