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.
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.
So far we've only seen K as an integer, how about when it isn't an integer, what will the correlation now be, here's a table with values of K when it isn't an integer. Value of K Roots of equation to 8 D.P -1/4 -1.469922111 -1/3 -1.434734764 -1/2 -1.362208742 1/2
However, my aim was to find the negative root, and I could not find it. Therefore, the Newton-Raphson method has failed to find that particular root, even though the starting value was close to it. Another example where the Newton-Raphson would not work is the function (x+21/2)1/2 ln (x+21/2)
bound is 2.8 and the function of the upper bound is 1.74() and they have no sign change due to it has an asymptote between 1 and 2 therefore there is no sign change and this method will fail when it is not continuous function.
By calculating the formula for the first two rows in Appendix 1 then dragging these formulae down the cells uses the formulae with the corresponding values to enable fast calculation of the area. By manipulating the formulae the volume of the other designs can be generated.
be a polynomial Then F(x) = Q(x) * divisor + remainder Sketching Graphs of y= 1/f(x) We can sketch a graph of y=1/f(x) if we already know the graph of y = f(x) Where y=f(x) crosses the x axis, y=1/f(x) has vertical asymptotes. 1/f(x)
3rd 18 From the table above, it shows there are 36 calculations for change of signs, 5 calculations for Newton Raphson and 18 calculations for x=g(x). To conclude, Newton Raphson method is the quickest in find the root, followed by x=g(x) method and the slowest is change of signs method.
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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