Formula for solving quadratic equations

A quadratic functions in the variable of x is an equation of the general form ax^2 + b x + c = 0, Where a, b, c are real numbers, a not equal to Zero. 2x^2 + x – 300 = 0 is a quadratic equation.

The simplest way to solve ax2 + bx + c = 0 for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the given quadratic equation is too messy, or it doesn't factor at all. Factoring may not always be successful; the Quadratic Formula always helps to find the solution.

The Quadratic equation form is ax2 + bx + c, where a, b, and c are just numbers; they are known as the numerical coefficients. The Formula is derived by the process of completing the square.

That is, ax2+ bx + c = 0, a not equal to Zero, is called the standard form of a quadratic function


Formula for Solving Quadratic Equation:


The solutions of any quadratic equation, ax2 + bx + c = 0 is given by the following formula, called the formula of quadratic equation:

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

For using the Quadratic Formula to work, the equation must be re arranged in the form (quadratic) = 0. Also, the 2a in the denominator of the Formula is underneath everything above in the numerator, not just the square root. And there 2a  is under there, not just a plain 2.


solving quadratic equation - Example problems:


solving quadratic equation -  problem 1:

x2  + 9x + 8  = 0

Solution:

Factoring method is used to solve for x,

x2 + 1x + 8x + 8 = 0

x(x + 1) + 8(x + 1) = 0

(x + 1) ( x + 8 )=0

X = -1, -8

The answers are -1,-8.

solving quatratic equation -  Problem 2:

Solve: x2 + 8x + 12 = 0.

Solution:

Factoring method is used to solve for x,

x2 + 5x + 3x + 12=0

x(x+5) + 3(x + 4) =0

(x+5)(x+3)=0

x = - 5 , - 3

The answers are  -5, -3.

solving quatratic equation -  Problem 4:

Solve 3x2 + 5x = -1 for x.

Solution: First find the standard form of

The equations and determine a, b,and c.

2x2 + 6x + 1 = 0

a = 2

b = 6

c = 1

Plug the values you found for

a, b, and c into the

Quadratic formula.

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

Perform any indicated operations.

X = ` ( -6 +- sqrt(6^2 - 8))/(4) `

X = `(-6 +- sqrt (28)) / 4`

The solutions are as follows:


X =`(-6 + sqrt (28)) / 4`           and X =  `(-6 - sqrt (28)) / 4`

sine rule formula

In the trigonometry, Sine rule formula (also called as sine formula or law of sines or sines law or sine rule) is the equation that can be used for relating the lengths of the sides of the arbitrary triangle to sines of the corresponding angle. According to the sine rule formula,

`a/sin A = b/sin B =c/sin C`

The sine rule formula in the trigonometry defines that the ratio of Sine of the angle with respect to its included sign and the vice versa depending on number of the properties in an obtuse triangle. If the obtuse triangle has two sides with given included angle as SSA(Side-Side-Angle) or ASA(Angle-Side-Angle) or AAS(Angle-Angle-Side). Then the Sine Rule Formula can be used to find the Angle of one of the sides.

a, b, and c are the lengths of sides of a given triangle, and A, B, and C are the opposite angles. Sometimes the sine rule formula is also stated as the reciprocal of this equation:

`sin A/a = sin B/b =sin C/c`

The sine rule formula is used to compute the remaining sides of the triangle when the two angles and one side is known. The technique is known as triangulation. It is used when two sides and one of the non-enclosed angle is known. In such cases, the sine rule formula gives two possible values for an enclosed angle.

Proof of the Sine rule formula


From the above triangle AXC,

`sin A=h/b`

`bsin A=h`

 From triangle XBC,

`"sin B `

`asin B=h`

Equating both the Equations we have

`h=bsin A=asin B`

So,

`b/sin B=a/sin A`

 A perpendicular from A to BC, we can show that

`b/sin B=c/sin C`

Hence we have the Sine Rule:

 `a/sin A = b/sin B =c/sin C`

The area of a triangle :

The Area of any triangle is `1/2 ab sinC` using the sine rule formula.

The Sine rule formula can be used if we don't know the height of a triangle (since we have to know the height for

`1/2 (base) * (height)` .


Examples using the sine rule formula


Q 1: Given side a = 10, side c = 14, and angle C = 30°

Sol: Using the Sine rule formula , we conclude that

`sin A / 20 = sin 30 /14`

`sin A = 1/2 * 20/14`

`A = sin ^-1 (1/7)`

A = 0.14°

2)  Applying the sine rule formula :

`17/sin 62 = 13/sin theta`

`17 sin theta = 13 sin 62`

`sin theta = (13 sin 62)/ 17`

`sin theta = (13 * 0.8829)/17`

`sin theta = 0.6752`

`theta = sin ^-1 (0.6752)`

? = 42.47°

Solving Online Rectangle Properties

A rectangle is any quadrilateral with four right angles. The term oblong is generally used to denote to a non-square rectangle. A rectangle with vertices WXYZ would be denoted as WXYZ. A so-called crossed rectangle is a crossed quadrilateral which consists of two opposite sides of a rectangle along with the two diagonal.





(source: Wikipedia)

Properties of Rectangle - Solve Online Rectangle Properties

Symmetric Properties:

cyclic               - The four corners touch the same circle
Equiangular  - The corner angles are equal (i.e.) 90 degrees
Isogonal        - All corners are present within the same symmetry orbit

Possess reflectional symmetry  and rotational symmetry

Rectangle - Rhombus Duality:

Dual polygon of a rectangle = A rhombus



Formula:

Area = l `xx ` b  sq.units
Perimeter = 2 ( l + b ) units
Length of diagonal = `sqrt(l^2+b^2)`
when l = b, the rectangle becomes square

Example Problems on Solve Online Rectangle Properties

Find the perimeter of the rectangle of sides 4m and 1m

Solution:

Step 1: Let, l = 4m, b = 1m

Step 2: Perimeter of the rectangle = 2 (l + b)

Step 3: Perimeter of the rectangle = 2 (4+ 1)

= 2 (5)

= 10m

Result: Therefore, Perimeter of the rectangle = 6m

2.  Find the area of the rectangle of sides 4m and 1m

Solution:

Step 1: Let, l = 4m, b = 1m

Step 2: Area of the rectangle = l `xx` b

Step 3: Area of the rectangle = 4 `xx` 1

= 4 m2

Result: Therefore, Area of the rectangle = 4 m2

3.  Find the length of a diagonal of  the rectangle of sides 1m and 4m

Solution:

Step 1: Let, l = 4m, b = 1m

Step 2: Length of a diagonal of the rectangle =   `sqrt(l^2+b^2)`

Step 3: Length of a diagonal of the rectangle = `sqrt(4^2+1^2)`

= 16.03 m2

Result: Therefore, Length of a diagonal of the rectangle = 16.03 m2

Practice problems on Solve online Rectangle Properties:

Find the perimeter of the rectangle of sides 2cm and 3 cm
Find the area of the rectangle of sides 2cm and 3cm
Find the length of a diagonal of  the rectangle of sides 2cm and 3cm

Solutions for practice problems on Solve online Rectangle Properties:

10 cm
6 cm 2
3.6cm

Power of a Point Theorem

In math, power of point theorem is a theorem related to circle. Power of point theorem tells the relationship between the intersecting lines of a circle. Base on the intersecting lines, we have three possibilities for power of point theorem. This article gives a clear explanation of power of point theorem with some example problems.

Explanation to Power of Point Theorem;

Three possibilities for power of point theorem:
Case: 1



In the above figure, there are two intersecting lines intersect the circle insides. Then the power of point theorem is AE . CE = BE. DE

Case 2:



In the above figure, one of the line is tangent to the given circle. Then the power of point theorem is, AB2 = BC . BD

Case 3:



In the above figure, there are two lines intersect outside of the circle. Then the power of point theorem is CB . CA = CD . CE

Special case of power of point theorem:



In the above figure, there are two tangent lines. Then the power of point theorem is PA = PC

Example Problems to Power of Point Theorem:

Example: 1

Determine the unknown value of the following figure using power of point theorem.



Solution:

Given:

CB = 2

AB = 4

CD = x

DE = 1

Step 1:

Here two lines are intersecting at outside. So, the power of point theorem is, CB . CA = CD . CE

Step 2:

CB . CA = CB . (CB + CA)

= 2 . (2 + 4)

=2 . 6

= 12

Step 2:

CD . CE = x . (CD + DE)

= x . (x + 1)

= x2 + x

Step 3:

CB . CA = CD . CE

12 = x2 + x

x2 + x - 12 = 0

(x + 4)(x - 3) = 0

x = -4, 3 (Discard the negative answer)

x = 3

Answer: 1

Example: 2

Determine the unknown value of the following figure using power of point theorem.



Solution:

Given:

EA = x

ED = 2

EC = 5

EB = 5

Step 1:

Here two lines are intersecting at inside. So, the power of point theorem is,AE . CE = BE. DE

Step 2:

x . 5 = 5 . 2

5x = 10

x = `10/5`

x = 2

Answer: x = 2

Example Problems to Power of Point Theorem:

Problem: 1

Determine the unknown value of the following figure using power of point theorem.



Answer: 4

Problem: 2

Determine the unknown value of the following figure using power of point theorem.



Answer: 6

Sine Calculator

In mathematics sine is based on trigonometry. The trigonometry is the branch of mathematics it   deals with dealings between the angles of a triangle and sides. The sin determines the sides of the quadrilateral and it provisions the length of the residual diagonal. It is nothing but reformulation of Ptolemy’s theorem. Let u see sin calculator.

Sin Basic Addition Proof:

Let us see sin addition calculator. It one of the basic operations Here we can see sin addition formula proof.


Ptolemy’s theorem:



Here    BC = 1 AB= cos (α)

AC = sin (α) BD= cos (β)

DC = sin (β) AD = sin (α + β)

Therefore sin (α + β) = sin (α) cos (β) + cos (α) sin (β)

Sin addition formula: sin (a + b) = sin (a) cos (b) + cos (a) sin (b)



Sin the area of a triangle is half the creation of the elevation and the base, it follows that the area of a triangle is half the creation of any of the two side times sin of the included angle. For the three triangles in the diagram we have,

2Ac= c (d cosb) sin a,

2Ad= d (c cos a) sin b,

2A= cd cos (a + b)

This after cancelling cd gives the addition formula for sin.  Following examples are showing Sin values in calculator.


Sine Calculator:

Let us see find sin value in calculator. Following statements are showing briefly explanation.



Example 1:

Find sin 6 value in calculator:

Solution:

Using sin table sin 6 = .1045

Example 2:

Find sin 6 value in calculator:

Solution:

Using sin table sin 45 = .7071

Definition of Cos

Cos is one of the trigonometric ratios. The six basic trigonometric ratios are sin, cos, tan, sec, cosec and cot. Of these sin and cos are most familiarly used. Trigonometric ratios involve angles in their standard positions. They can be defined using a Right triangle as well as by a Circle.

Let us see the Definitions of Cos

Definition 1: using a right triangle



The sides of a right triangle are

1. Hypotenuse : The side opposite to the right angle (`90^o` ), in this case side `BC`
2. Opposite side: The side opposite to the angle of consideration (angle `theta` ), in this case side `AB`
3. Adjacent side: The side adjacent to the angle of consideration(angle `theta` ), here it is side `AC`

Definition of cos : Cosine is defined as the ratio of adjacent side to the hypotenuse

So, `cos(theta) = (adjacent side) / (hypoten use)`

`= (AC)/(BC)`

Here cos is defined for angles  `0^o< theta < 90^o`

Note: here ratio does not depend on the lengths of a particular triangle, it is same for all the right triangles which contain the  angle `theta` . Because all such right triangles will be simillar.

Definition 2: Using  Circle

Consider a circle with center  ‘`O` ’ at origin and radius ‘`r` ’ in the rectangular cartesian system. Let `theta` be any angle in standard position such that its terminal ray intersects the circle in point P(x,y).



From the figure

We have `OP = r`

And `x^2+y^2 = r^2`

Now cosine is defined as   `cos(theta)=x/r`

Here cos is defined for angles `0^o < theta < 360^o`

Examples for Definition of Cos :

For  a triangle  with sides AB =3, BC=5 ,AC=4 Right angles at ‘`A` ’


Calculate  `CosB` and `CosC`

Solution:

For calculating `cosB` :



Adjacent side is AB =3 , opposite side is AC =4, Hypotenuse is BC= 5

So `cos B = (adjacent side)/( hypoten use)`

`=(AB)/(BC)`

` = 3/5`

For calculating `Cos C` :



Adjacent side is AC =4, Opposite side is AB =3, Hypotenuse BC =5

So, `cos C = (adjacent side)/ (hypoten use)`

` = (AC)/( BC)`

` = 4/5`

Learning Natural Tangent Tables

The branch of math that deals with the relations of the sides and angles of triangles is known as trigonometry. Which the methods of reducing from a certain given parts to other required parts, and also in the general relations which exist between the trigonometrical functions of arcs or angles. The trigonometry ratio table is used to find the values of the trigonometry functions. Here we are going to see about learning of natural tangent tables:

The natural tangent of an angle is the ratio, the length of the opposite side to the length of the adjacent side. In our case

Tan A = `(opposite side) / (adjacent side)`

The natural tangent angle table will be very useful for finding the value of tangent angles. Learning of natural tangent tables will be very helpful when we are using the problems.


Learning Natural Tangent Tables:

Angle   Tangent(A)

Angle    Tangent (A)

0         0.000 1         0.017 2         0.035 3         0.052 4         0.070 5         0.087 6         0.105 7         0.123 8         0.141 9         0.158 10        0.176 11        0.194 12        0.213 13        0.231 14        0.249 15        0.268 16        0.287 17        0.306 18        0.325 19        0.344 20        0.364 21        0.384 22        0.404 23        0.424 24        0.445 25        0.466 26        0.488 27        0.510 28        0.532 29        0.554 30        0.577 31        0.601 32        0.625 33        0.649 34        0.675 35        0.700 36        0.727 37        0.754 38        0.781 39        0.810 40        0.839 41        0.869 42        0.900 43        0.933 44        0.966 45        1.000

45        1.000 46        1.036 47        1.072 48        1.111 49        1.150 50        1.192 51        1.235 52        1.280 53        1.327 54        1.376 55        1.428 56        1.483 57        1.540 58        1.600 59        1.664 60        1.732 61        1.804 62        1.881 63        1.963 64        2.050 65        2.145 66        2.246 67        2.356 68        2.475 69        2.605 70        2.747 71        2.904 72        3.078 73        3.271 74        3.487 75        3.732 76        4.011 77        4.331 78        4.705 79        5.145 80        5.671 81        6.314 82        7.115 83        8.144 84        9.514 85        11.430 86        14.301 87        19.081 88        28.636 89

Examples for Learning Natural Tangent Tables:

Find the values of the given natural tangent angles using natural tangent table.

1) tangent 1   =      0.017

2) tangent 48   =   1.1114

3) tangent 9   =     0.158

4) tangent 56  =     1.483

5) tangent 16   =     0.287

6) tangent 64   =     2.050

7) tangent 25    =    0.466

8) tangent 78    =    4.705

9) tangent 44   =    0.966

10) tangent 88   =    28.636