Uses of a Quadratic Equation

Introduction to Uses of a quadratic equation:

In this article we are going to discuss the uses of quadratic equation solving. The general form of quadratic equation is ax2 +bx +c, the value of is not equal to zero. The uses graph of this function is a parabola have vertical axis. For example f(x) =2x2 + x – 30 = 0 is a quadratic equation.

Uses of a Quadratic Equation –quadratic Equation Formula:

Formula for finding the roots of the quadratic equation is,

` -b +- sqrt(b^2-4ac)/(2a)`

Mathematic quadratic equation example problems are given below.

Uses of a Quadratic Equation -example Problems:

Example 1:

f(x)=x2+8x+16 = 0 solve by uses of factorize method for quadratic equation.

Solution:

The quadratic equation solving by uses of factoring method, to split the middle part (8x) into two parts so that the product of their co-efficient is equivalent to the constant part (16).

Like 8x = (4x) and (4x)

So, 4x + 4x = 8x and

4 * 4 (coefficients of 4x and 4x) = 16 (constant term)

So, now the equation becomes

x2 + 4x + 4x + 16 = 0

Here ‘x’ in 1’st part and 4 in last two parts are equal, by taking both part of equal out, we get

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

Now we have (x+4) in common

(x+4) (x+4)  = 0

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

x= -4 (or) x= -4

The solution of quadratic equation x is -4 (or) -4

Example 2:

Solve 3x2 - 6x = -2 for x, uses of quadratic equation formula,

Solution:

Using standard form of ax2+bx+c=0

3x2 - 6x + 2 = 0

a =3

b = -6

c = 2 Plugging the values you found for a, b, c in the quadratic equation formula.

x =` -b +- sqrt(b^2-4ac)/(2a)`

x = `6 +- sqrt(36-24)/ 6`

x = `6 +- sqrt(12)/6`

x = `6 +- 2 sqrt(3)/6`

The solutions are as follows:

The solution of quadratic equation is x = 6 + 2 `sqrt(3)/6` and 6 - 2 `sqrt(3)/6`

Length of Right Triangle Sides

Introduction to length of right triangle sides:

In a triangle the one angle having 90 degree and the sum of other two angles is 90 degree means, that triangle is known as right angle triangle. And right triangle satisfies the Pythagoras theorem. The sides of the right triangle is in the manner of one largest side known as hypotenuse  and two legs called adjacent side and opposite side .



Explanation of Length of Right Triangle Sides:

Types of Right triangle:

General Right triangle
Isosceles Right triangle
30-60-90 Right triangle.
General Right triangle:
General Right triangle having one angle 90 degree and other two angles can be any measure that total yields 90 degree.

Isosceles Right triangle:

In Isosceles Right triangle the other two angles are in the measure of 45 degree each. In isosceles right triangle the sides are in the ratio of 1:1: `sqrt(2)` (adjacent side: opposite side: hypotenuse)

30-60-90 Right triangle:

As the name itself noted that this kind of right triangle having one angle is 90 degree mandatory and in other angles one is 30 degree and the other one is 60 degree. The ratios of the sides are 1:`sqrt(3)` :2 (adjacent side: opposite side: hypotenuse).





Pythagoras Theorem for the Length of Right Triangle Sides:

The Pythagoras theorem states that,

In a right angle triangle the total sum of the squares of the two sides (adjacent and opposite) are equal to the square of the longest side (hypotenuse) understanding hard math problems for 9th graders is always challenging for me but thanks to all math help websites to help me out.

Let a, b, c are the three sides of a right triangle where a, b are two legs and c is the longest side . Then the formula of Pythagoras theorem is,

c2= a2+b2

c= `sqrt(a^2+b^2)`

Examples on Length of Right Triangle Sides

Ex:1 In a right triangle the length of the adjacent side is 12cm and opposite side is 14cm. Find the length of the hypotenuse.

Sol:   Let a= 12cm, b=14cm and c= hypotenuse

By Pythagoras theorem,

c= `sqrt(a^2+b^2)`

= `sqrt(12^2+14^2)`

= `sqrt(144+196)`

= `sqrt(340)`

= 18.43

Hence the length of the longest side is 18.43 .

Ex:2 The length of the adjacent side and opposite side is 15cm in a right triangle. Find the length of the longest side.

Sol:  Let a= b=15 cm and c= hypotenuse

By Pythagoras theorem,

c= `sqrt(a^2+b^2)`

=  `sqrt(15^2+15^2)`

= `sqrt(225+225)`

=`sqrt(450)`

=21.21

Hence the length of the hypotenuse is 21.21

Summary on Quadratic Functions

Introduction on summary on quadratic functions

Quadratic functions are polynomial functions containing a quadratic expression. A quadratic expression is an algebraic expression of the degree 2.
A quadratic function can be written in different forms,

Standard form

A quadratic function of the form of `f(x) = ax^2 + bx + c` is in the standard form. In this form, a, b, and c are real numbers and `a != 0` . Furthermore, a, b and c are called the Coefficient of x square, Coefficient of x and Constant term respectively. The vertex of the parabola formed by graphing a quadratic function can be obtained by the formula `((-b)/(2a), (-1(b^2 - 4ac))/(4a))` .

Summary on Quadratic Functions-2 Forms

Intercept form

A quadratic function of the form of `f(x) = k(x - a)(x - b)` , that is, of the form of the product of two linear expressions, is in the intercept form. To graph a quadratic function, this form is converted into the standard form by expanding (multiplying the two linear expressions and number 'k').

Vertex form

A quadratic function of the form of `f(x) = a(x - h)^2 + k` is in the vertex form. The vertex of the parabola formed by graphing a quadratic function is given by `(h, k)` .

Graphing a Quadratic Function -summary on Quadratic Functions

The graph of a quadratic function gets the shape of a parabola, which is a conic section. A conic section is the surface obtained by intersecting a cone or a conical figure.
Let us learn the method of graphing a quadratic function by graphing the function `f(x) = x^2 - 5x + 6`

On comparing the given function with `f(x) = ax^2 + bx + c` , we get `a = 1` , `b = -5` and `c = 6` . Vertex of the parabola to be formed is given by `((-b)/(2a), (-1(b^2 - 4ac))/(4a))` .
Thus, vertex = `((5/2), (-1)/(4))`
First plot the vertex on the graph.

Since a parabola is a curve, we need the coordinates of many points lying on it. To obain these points, choose different (at least 4) values for the variable `x` , and plug in those values in the function to get the corresponding values of the function. For the function `f(x) = x^2 - 5x + 6` , we choose the values `x = 1` , `0` , `-1` , and `2` and obtain the following pairs of coordinates:-
Input value (x)    Output value (f(x))
0    6
1    2
-1    12
2
0

All the above pairs of coordinates are of the points lying on the parabola. Graph them, and then join them to form a parabola. This parabola is the graph of the function `f(x) = x^2 - 5x + 6` . It will look as follows:-

PreCalculus Problem Solver

Introduction to Precalculus Problem Solver:

The Precalculus is one of the foundation classes of mathematics. The topics involved in Precalculus test are real  complex numbers, binomial theorem, composite & polynomial functions, vectors, parametric equations, polar coordinates, matrices, rational functions, solving inequalities & equations and trigonometry. In this article we shall discuss about the examples involved in Precalculus Problem Solver. The following are the examples involved in Precalculus Problem Solver.

Precalculus Problem Solver:

Example 1:  Evaluate the distance between the points (2, 5) and (-1, 2) using distance formula.

Solution :   Given (x1, y1) = (2, 5)

(x2, y2) = (-1, 2)

The distance formula D = `sqrt((x2-x1)^2+(y2-y1)^2)`

= `sqrt((-1-2)^2+(2-5)^2)`

=  `sqrt((-3)^2+(-3)^2)`

= `sqrt(9 + 9)`

= `sqrt(18)`

Example 2:  Express the real and imaginary parts for 19 - i `sqrt(5) `

Solution:        Let z= 19 - i `sqrt(5) `      

Re(z) = 19    Im(z) = `sqrt(5)`

Example 3:   Find the vertex of the parabola y = 2x2 – 16x + 6

Solution:  Given:  y = 2x2 – 16x + 6

We know that x = -`(b)/(2a)` ,

Here a = 2, b = -16

So that,   X =` -b/(2a)` = `-((-16))/(2*2)` = 4

And then y = 2(22) – 16(2) + 6 = 8 – 32 + 6 = -18

So, x = 4 and y = -18.

Example 4:   If f(x) = 7x-1, find f-1(y)?

Solution:Let y = 7x - 1

The given equation can be rewritten as

y + 1 =7x

=> x = y + 1
7

Therefore the f-1(y) = y + 1
7

Example 5: Write the given number in complex form `sqrt(-44)`

Solution: `sqrt(-44)` = `sqrt((-1)(44))`

= `sqrt(-1)` * `sqrt(44)`

= i `sqrt(44)`

Having problem with solve math problem for me keep reading my upcoming posts, i will try to help you.

Precalculus Problem Solver:

Problem 1:Write the given number in complex form `sqrt(-175)`

Answer: = i `sqrt(175)`

Problem 2:Write the real and imaginary parts for 22 - i `sqrt(7) `

Answer:   Re(z) = 22    Im(z) =`sqrt(7)`

Problem 3:   If f(x) = x-9, find f-1(y)?

Answer: y + 9

Problem 4: Evaluate distance between the points (4, 5) and (-1, 2) using distance formula.

Answer: D= `sqrt(34)`

Steps for Rational Expressions

Introduction to steps for rational expressions:

Rational expressions are defined as the fractional number but instead of number the polynomials are present. By using the rational expressions we can able to do all the arithmetic operations that we performing in mathematics. For example, the rational expressions are in the form of `A/B` , where the numerator and denominator are called the polynomials terms.

Steps for Rational Expressions

Steps of rational expressions are follows,

By using the rational expression, we can able to do all the arithmetic operations.
Certain rules are followed for each of the operations performed for rational expressions.
The operation performed on the rational expressions are,
Addition
Subtraction
Division
Multiplication
Cancellation
Reciprocals

Example Problem for Steps of Rational Expressions

Problem 1: Add the following rational expressions, `(2x)/(3x^2)`  +  `(3x)/(3x^2)`.

Solution:

Step 1: Write the given rational expressions, we get,

`(2x)/(3x^2)`  +  `(3x)/(3x^2)`

Step 2: Check the denominator same or not,

Step 3: Add the numerator terms, we get,

`(2x + 3x)/(3x^2)` 

Step 4: Solve the obtained result, we get,

`(5x)/(3x^2)`

This is the required answer for the rational expressions.

Problem 2: Add the following rational expressions, `(4x)/(2y^2)`  +  `(4x)/(2y^2)`.

Solution:

Step 1: Write the given rational expressions, we get,

`(4x)/(2y^2)`  +  `(4x)/(2y^2)`

Step 2: Check the denominator same or not,

Step 3: Add the numerator terms, we get,

`(4x + 4x)/(2y^2)` 

Step 4: Solve the obtained result, we get,

`(8x)/(2x^2)`

This is the required answer for the rational expressions.

Problem 3: Add the following rational expressions, `(y^2)/(8x^2)`  +  `(4y^2)/(8x^2)`.

Solution:

Step 1: Write the given rational expressions, we get,

`(y^2)/(8x^2)`  +  `(4y^2)/(8x^2)`

Step 2: Check the denominator same or not,

Step 3: Add the numerator terms, we get,

`(y^2 + 4y^2)/(8x^2)` 

Step 4: Solve the obtained result, we get,

`(5y^2)/(8x^2)`

This is the required answer for the rational expressions.

Practice Problem for Steps of Rational Expressions

Problem 1: Add the following rational expressions, `(x)/(5x^2)`  +  `(x)/(5x^2)`.

Answer: The answer for the above problem is `(2x)/(5x^2)`

Problem 2: Add the following rational expressions, `(6x)/(2y^2)`  +  `(8x)/(2y^2)`.

Answer: The answer for the above problem is  `(14x)/(2y^2)`

Integral Change of Variable

Introduction to integral change of variable:

In this article, we study about integral change of variable and their example problems. Integral means finding the anti derivative of the function. Integral can be classified as two types. They are indefinte integral and definite integral. Improper integral also used in change of variable. Change of variable means change the variable of the given integral function. If the variable of the changed means, we also change the limit values of the given integral function. Change of variable is mainly used in definite integral problems.

Example Problems for Integral Change of Variable

Integral change of variable example problem 1:

Find the integral of the given function

` int_0^4 sin(4x)dx`

Solution:

Given function is `int_0^4 sin(4x)dx`

Using the change of variable method,

Take u = 4x

Therefore, for finding dx differentiate the u value, we get

du = 4 dx

Rearrange the above value, we get

dx = `(du)/(4)`

The limits are also changed,

When x = 0, u = 0

When x = 4, u = 16

The given integral function can be written as,

`int_0^4 sin(4x) dx` = `int_0^16 sinu (du)/(4)`

After rearranging the above function, we get

= `(1/4) int_0^16 sinu du`

Integrating the above function, we get

= `(1/4) [- cosu]^16_0`

Substituting the limit values, we get

= `(1/4) ((- cos 16) - (- cos0))`

= `(1/4) ((- 0.96) - (- 1))`

After simplifying, we get

= 0.01

Answer:

The final answer of the given function is 0.01

Integral Change of Variable Example Problem 2:

Find the integral of the given function

`int_0^3 ((1)/(2x + 3))dx`

Solution:

Given function is `int_0^3 ((1)/(2x + 3))dx`

Using the change of variable method,

Take u = (2x + 3)

Therefore, for finding dx differentiate the u value, we get

du = 2 dx

Rearrange the above value, we get

dx = `((du)/(2))`

The limits are also changed,

When x = 0, u = 3

When x = 3, u = 9

The given integral function can be written as,

`int_0^3 ((1)/(2x + 3))dx` = `int_3^9 ((1)/(u))(du)/(2)`

After rearranging the above function, we get

= `(1/2) int_3^9 (1/u) du`

Integrating the above function, we get

= `(1/2) [logu]^9_3`

Substituting the limit values, we get

= `(1/2) ((log 9) - (log3))`

= `(1/2) ((0.95) - (0.47))`

After simplifying, we get

= 0.24

Answer:

The final answer of the given function is 0.24

My Previous Blog :- http://advancemath.blogspot.in/2012/08/complex-fraction-calculator.html

Area of a Triangle by Heron's Formula

Introduction to area of a triangle by heron's formula:    
Heron formula defined by heron of alexandria and in METRICA book in 60 A.D.It is found by chinese.They published in 1247 A.D.Area of a triangle is calculated by using this formula.The length of sides are b,c,d and it is used in area of triangle A.It is most widely used in geometry.The formula for finding area of triangle is

A = vS(S-b)(S-c)(S-d) , where  S is the semiperimeter of the triangle.

S= (b+c+d) / 2.

Area of a Triangle by Heron's Formula-example

Problem 1: Find the area of below triangle:



Answer: Using heron's formula,

p= 34.5+16.4+19.6 / 2 = 35.26.

Area = `sqrt(35.26*0.76*18.86*15.66)`

= 89.75.

Therefore, the area of triangle is 89.75.

I am planning to write more post on examples of substitution method, prime numbers under 1000. Keep checking my blog.

Problem 2: Find the area of triangle from given value.

A=34.5, B= 28.8, C= 12.0

Answer: Using heron's theorem,

p = 34.5+28.8+12.0 / 2 = 37.64.

Area = `sqrt(37.64*3.14*8.84*25.64)`

= 164.

Therefore the area of triangle is 164.

Excercise for heron's theorem:

Problem: Find the area of triangle from A= 15.7, B= 34.5 and C =19.5 using heron's theorem.

Answer: Area of triangle =62.35.