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.
Therefore, the higher the R^2 (0?R^2?1), the closer the estimated regression equation fits the sample data. In this case, R^2 is 0.99624, which very close to unity. It indicates that the absolute equation fits the data very well according to R^2.
Some may use symmetry, some just trial and improvement, others may think about patterns of numbers. Extension work: Can the mean be 5? Can you get the mean to be between 6 and 7? * Obtain a mean of 5 and mode of 4 * Obtain a mean of 6 and mode of 3.
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).
27=a5+b (2) 9=a (2)-(1) Substitute 'a' back into (1) 18=9x4+b 18=36+b b= -18 Substitute 'a' and 'b' back into second number equation S=9w-18 So we have the equations for the first and second numbers of the table. So now we put them together. The first number equation is solved by 4W-9 and then to that you had to times
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
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.
Comparisons Now, I am going to compare all the 3 methods I have used in the coursework, including change of signs method, Newton Raphson method and x=g(x) method. In order to do so, I will first use the equation which was used in change of signs, x³+3x²–3=0, and use the Newton Raphson method and x=g(x)
Over 160,000 pieces of student written work
Annotated by experienced teachers
Ideas and feedback to improve your own work
Want to read the rest?
Sign up to view the whole essay and download the PDF for anytime access on your computer, tablet or smartphone.