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Bisection Method (real root of equation)
- find a and b via f(x) = ... (one +ve and -ve) must be latest following the f(num) which get -ve or +ve would be a or b ^858dca
- after that make table a | b | f(a) | f(b) | x = a+b/2 | f(x)
- Concept if f(x) is -ve, x=a & f(x) = f(a) and if f(x) is +ve, x=b & f(x) = f(b)
- if question says up to n decimal then when that decimal repeat then it is real root of the given eqn otherwise do up to 3 decimals for x
Newton Ralph Method
- first find a & b with table method [[#^858dca|see this]]
- do let initial approximation (x0) = 2.5 (any number between a & b)
- now then get derivative of f(x) is f'(x)
- nr formula : $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
- now get x1 with the formula when n is 0 becomes x0 the initial value 2.5
- keep on going till x_n until the 3 decimal or given decimal answer match
Secant Method
- first find a & b with table method [[#^858dca|see this]]
- secant-method forumula : $$x_n = \frac{a.f(b) - b.f(a)}{f(b) - f(a)}$$
- now we make table for it and keep calculating until the same decimal match : iteration(n) | a | b | f(a) | f(b) | X_n | F(x_n)
- here X_n is got by secant formula and all other for first comes from a & b then their f(x) value and f(x_n)

TIP TO FIND EXACT ANSWER FOR BISECTION AND RALPH

To find the exact answer guess get value a & b from the table method and the equivalent f(x) of those values and which ever value is nearer to 0 a or b that becomes Ans: and this is formula On table method before make sure it is in f(x) = 0 standard form. $$\text{Ans} - \frac{f(\text{Ans})}{f'(\text{Ans})}$$

here the $${f'(\text{Ans})}$$ means derivative.

Term in f(x)	Derivative in f′(x)	Why?
$x^4$	$4x^3$	Power Rule
$5x^3$	$15x^2$	$5 \times 3 = 15$
$x$	$1$	$x^1$ becomes $1x^0$
$7$ (any number)	$0$	Numbers don't change slope
$-2x^2$	$-4x$	$-2 \times 2 = -4$