Functions Solving Online

Introduction to Functions solving online:

The mathematical idea of a function expresses dependence between two quantities, one of the produced is   independent variable, argument of the function, or it’s "input" and another one of the produced is dependent variable, value of the function, or "output". A function expression is associates a unique output with all input element from a fixed set, such as the real numbers.

The function of expression is f(x) =ax^2 + b x + c


Explain about Functions Solving online using quadratic functions


Quadratic function in online is helpful to know more about the simplest way to solving online functions which under  "ax^2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. While factoring could not always be successful, in online the Quadratic Formula can always find the solution by using the .

The general form is,

ax^2 + bx + c = 0

Where, x represents the variable, and a, b, and c, constants, with a ? 0. (If a = 0, the equation becomes a linear equation.) .

Is this topic Implicit Function hard for you? Watch out for my coming posts.


Examples of Functions solving online using Quadratic functions


Using Quadratic functions solving some example problems are given below,

Ex 1 :   solving qudratic functions f(x) = x^2+6x+9 = 0

Sol :  To factorize the quadratic function, split the middle term (6x) into two terms so that the product of their coefficients is equal to the constant term (9).

Like 6x = (3x) and (3x)

So, 3x + 3x = 6x and

3 * 3 (coefficients of 3x and 3x) = 9 (constant term)

So, now the function becomes

x^2 + 3x + 3x + 6 = 0

Here ‘x’ in first term and 3 in last two terms commonly, by taking both of them commonly out, we get

x(x+3) + 3(x+3)  = 0

Now (x+3) in common

(x+3) (x+3)  = 0

Now x+3 = 0 or x+3 = 0

Ex 2 :  solving qudratic function f(x) =x^2+6x+8 = 0

Sol :    To factorize the quadratic function, we have to split the middle term (6x) into two terms so that the product of their coefficients is equal to the constant term (8).

Like 5x = (2x) and (4x)

So, 2x + 4x = 6x and

2 * 4 (coefficients of 2x and 4x) = 8 (constant term)

So, now the function becomes

x^2 + 2x + 4x + 8 = 0

Here ‘x’ in first term and 4 in last two terms commonly, by taking both of them commonly out, we get

x(x+2) + 4(x+2) = 0

Now we have (x+2) in common,

(x+2) (x+4)  = 0

Now x+2 = 0 or x+4 = 0