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, my prediction was correct. y=x^4 x y second point x second point y gradient 3 81 3.1 92.3521 113.521 3 81 3.01 82.08541201 108.5412 3 81 3.001 81.10805401 108.05401 4 256 4.1 282.5761 265.761 4 256 4.01 258.569616 256.9616 4 256 4.001 256.256096 256.09602 2 16 2.1 19.4481 34.481
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.
To verify this I am going to draw another graph to show that this is correct. So we can see here the graph definitely shows that the largest volume possible is 16 and that the length of the square cutting for this would be 1.
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.
This is because the mid-point overestimates whilst the trapezium rule underestimates. Therefore, by using both methods we can confine the exact value between the two different polynomials. This means by obtaining values of Tn and Mn where n is a large number, we can find an accurate approximation to the
making sure that no errors is made or I will get the wrong answer. Using a calculator, for example, I might press in 1.51218 instead of 1.51213 and this will give me the wrong answer. I think that Rearrangement Method should be the fastest of all three without the aid
I will use one root a as 0 and the other root b as ?/4. Bearing this in mind we can substitute the values into the formula: The formula is the same as the secant method however remember the false position method roots approximation must support the sign change argument.
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.
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