Quadratic Equations Activity

An equation with one variable, in which  the highest power of the variable is two is called a quadratic equation.

For example, ( i ) 3x2 + 5 x - 8 = 0

(ii) 2y2 - 48 = 0

(iii) 6x2 + 5x = 0

(iv ) y2 = 4 are all quadratic equations. Let us do some activity or problems using quadratic equations.

Activities of Quadratic Equations:

Activity 1 :

Solve the quadratic equation 2x2 - 7x = 39

2x2 - 7x - 39 - 0

`=>`              2x2  - 13x + 6x - 39 = 0  ( factorising the left hand side )

`=>`              x ( 2x - 13 ) + 3 (2 x - 13 ) = 0

`=>`             ( 2x - 13 ) ( x + 3 ) = 0

2x - 13 = 0 or x + 3 = 0

`rArr`              x  =  `13/2`   or  x  =  -3

Hence the quadratic equation is solved by factorisation method.

Activity 2 :

Find the quadratic equation whose solution set is { -2, 3 }

Since solution set is { -2, 3}

we have x = -2 or x = 3

x + 2 = 0 or x - 3 = 0

`rArr`           ( x + 2 ) ( x - 3 ) = 0

`rArr`           x2 - 3x + 2x - 6 = 0

`rArr`           x2 - x - 6 = 0 is the required quadratic equation.

Activity of Quadratic Equations(continued):

Activity 3 :

Solve the quadratic equation 5x2 - 2x - 3 = 0 using the formula.

The roots of the standard quadratic equation ax2 + bx + c = 0 where a`!=` 0, are given by the formula

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

Comparing 5x2 - 2x - 3 = 0 with ax2 + bx + c = 0 we get a = 5, b = -2 and c = -3.

so, x = `(2 stackrel(+)(-) sqrt((-2)^2 - 4. 5. (-3))/(2.5))`

=  `(2 stackrel( +)(-) sqrt ( 64)) / ( 10)`

= `(2 stackrel(+)(-) 8)/10`

= `(2-8)/10` = 1 and `-3/5`

Hence 1 and `(-3)/5` are the roots of the given quadratic equation.

Activity 4 :

Solve the equation 2x4 - 5x2 + 3 + 0 which is reducible to quadratic equation.

Let x2 = y

Then, 2x4 - 5x2 + 3 = 0   `rArr`   2y2 - 5y + 3 + 0

`rArr`   ( y - 1 ) ( 2y - 3 ) = 0

`rArr`   y = 1 or   y   = `3/2`

When y = 1, x2 = 1 `rArr` x = 1 or -1

When y =  `3/2`   x2 = `3/2` `rArr` x = `sqrt(3/2)`   or `-sqrt(3/2)`

Hence the fourth degree equation is solved using the quadratic equation technique.

Slope Ratio Calculator

The slope is defined as ratio of change of x axis to change of y axis. The slope intercept form is y = mx + b. Where m is slope and b is y intercept.

Slope formula is (m) =`"vertical" /"horizontal"`.

The slope form (m) is = `(y_2 - y_1)/(x_2 - x_1)` = `"rise"/"run"` .

We will learn about the slope ratio calculator example problems and practice problems are given below.

Example Problems for Slope Ratio Calculator:

Slope ratio calculator:





Example problem 1:

Find the slope ratio of a line, which contains two points A (0, 1), B (10, 2).

Solution:



The slope of a line which contains two points (x1, y1) and (x2, y2) is given by,

Here, x1 = 0, x2 = 10, y1 = 1, y2 = 2.

Slope of the line, m = `(y_2 - y_1)/(x_2 - x_1)`

= `(2 - 1)/(10 - 0)`

After simplify this, we get

= `1/(10)`

Slope of the line (m) = `1/(10)`

So, the slope ratio of a line, which contains two points A (0, 1), B (10, 2) is `1/(10)` = 0.1



Example problem 2:

Find the slope ratio of a line, which contains the two points A (-13, 10), B (2, -20).



Solution:

The slope of a line which contains two points (x1, y1) and (x2, y2) is given by,

Here, x1 = -13, x2 = 2, y1 = 10, y2 = -20.

Slope of the line, m = ` (y_2 - y_1)/(x_2 - x_1)`

=  `(-20 - 10)/(2 + 13)`

After simplify this, we get

= ` (-30)/(15)`

Slope of the line (m) = -2

So, the slope ratio of a line, which contains the two points A (-13, 10), B (2, -20) is -2

These examples problem are helpful to study of slope ratio calculator.

Practice Problems for Slope Ratio Calculator:

Practice problem 1:

Find the slope of a line, which contains two points A (9, 0), B (0, 3).

Answer: Slope (m) = -0.3333

Practice problem 2:

Find the slope of a line, which contains two points A (10, 200), B (300, 20).

Answer: Slope (m) = -0.620

Algebra Probability Help

Mathematical numbers are studied by algebra. Algebra is also used to learn the polynomials and the equations etc. Probability is a method of state knowledge or principle that an occurrence will happen. In mathematics the idea has been given a correct sense in probability theory, that is used widely in such areas of learn as mathematics, finance, statistics etc. Here we will see the examples and solved with the help of algebra probability.

Example Problems for Algebra

1) What is the multiplication of following two numbers with the help of algebra? 115*118

Solution

115*118 = (100+15)*(100+18)

= (100)2+(100*18)+(15*100)+(15*18)

=10000+1800+1500+270

=13570.

2) 18x+12y+12x+8a. Simplify the given equation in algebra.

Solution

The given equation is 18x+12y+12x+8a

There are two related groups are available. So connect the groups.

The new equation is,

(18x+12x)+12y+8a

Add the numbers inside the bracket. We get 30x+12y+8a.

Assemble the numbers and we get the correct format.

=8a+30x+12y.

Example for Probability

There are 45 things are available in a shop. In those things, 16 are the books, 12 are the bags and 17 are the caps. What is the probability for the following outcomes?

i) Select the books

ii) Select the bags

iii) Select the caps.

Solution:

Total number of things n(S) =45

Number of books n (A) =16

Number of bags n (B) =12

Number of caps n(C) =17

i) Assume P(A) is the probability for select the books.

P(A)=`(n(A))/(n(S))`

=`(16)/(45)` .

ii) Assume P(B) is the probability for select the bags.

P(B) =`(n(B))/(n(S))`

= `(12)/(45)`

=`(4)/(15)` .

iii) Assume P(C) is the probability for select the caps.

P(C)=`(n(C))/(n(S))`

=`(17)/(45)` .

Practice Problems

Practice problem 1

What is the product of the following numbers in algebra ? 111*128

Answer:

101*108=14208.

Practice problem 2

8x+14y+12a+12y. Simplify the terms with the help of algebra.

Answer:

12a+8x+26y.

Practice problem 3

Jenifer has the 6 papers, Stephen has 7 papers. What is the probability for select the Jenifer’s papers.

Answer:

Probability for select the Jenifer’s papers=6/13.

These algebra and probability problems are helping to study these concepts

Trigonometry Sine Function

In trigonometry one of the ratios is the sine function. It is defined by sinx = (Opposite side/Hypotenuse) = (Perpendicular/Hypotenuse)

This trigonometry sine function has applications on solving some practical problems in finding the height of the wall, length of a ladder leaning on the wall and the distance between the wall and the foot of the ladder.

Let us learn some sine values for standard angles.





The above table will help us in solving problems involving sine functions.  Now let us solve few problems on the topic trigonometry sine function.

Example Problems on Trigonometry Sine Function

Ex 1: From the below diagram, find the value of x using sine function.

Sol: Sin30 = `x/12`

This is implies, x = 12 `xx` sin30

                            = 12 `xx` `(1/2)` [table value for sin30 = `1/2` ]

                            = 6 cm.

Therefore, the value of x = 6cm.

Ex 2: From the below diagram, find the value of x using sine function.

Sol: Sin30 = `3/x`

               x = `3/sin30`

                  = `3/(1/2)`

                  = 3 `xx` 2 = 6cm.

Therefore, the value of x = 6cm.

Ex 3: If 2 sinA = 1, what is the value of A?

Sol: Given: 2sinA = 1

                     SinA = `1/2`

This implies that the value of A = `30^0` .     [Table value Sin30 = `1/2` ]

Ex 4: Simplify: sin60 + sin30.

Sol: sin60 + sin30 = `sqrt(3)/2` + `(1/2)`

                              = `(((sqrt(3)) + 1)/2)` .            [Table value Sin60 = `sqrt(3)/2` ]

Ex 5: In the triangle, find the value of x^0.

Sol: We know that Sin`x^0` = `sqrt (3)/2`

This implies `x^0` = `60^0` [Table value, sin60 = `sqrt (3)/2` ]

Ex 6: If 4sin²x – 3 = 0, x is an acute angle, find (i) sinx (ii) x.

Sol: Given: 4sin²x – 3 = 0

                     Sin²x = `(3/4)`

                     Sinx = `+-` `sqrt(3)/2`

(ii) Since, sinx = `+-` `sqrt(3)/2` ,

As per the table value, x =  `60^0` .

Therefore, x = `+-` `60^0` .

Ex 7: Solve for x:

Sin(x + 10) = ½

Sol: Since Sin(x + 10) = `(1/2)`

                        X + 10 = `30^0`

Therefore, x = 30 – 10

                     = `20^0` .

Therefore, the value of x = `20^0` .

Practice Problems on Trigonometry Sine Function

1. Solve for x: Sin² x + sin²30 = 1.

[Answer: x = `+-` `60^0` ]

2. Find the acute angle A and B, if Sin (A+B) = 1 and Sin(A – B) = `(1/2)`

[Answer: A = `60^0` , B =` 30^0` ].

Areas of Combinations of Plane Figures

Areas of Combinations of plane figure:

Areas of Combinations of plane figure is the process of calculating the areas of different combinations of figures. these types of figures  We come across in our daily life and also in the form of various interesting designs.

Flower beds, drain covers, window designs, We come across, designs   on th etable covers, are some of such examples.We illustrate the  process of calculating  areas of these  figures through some examples.

The following   examples are combined with some plane figures.

Areas of Combinations of Plane Figure Problems:

Example:

Two circular flower beds have been shown on two sides of  a square lawn ABCD  of side  50m.If the center of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.





Solution:

Area of  the square lawn ABCD = 50 x 50 m2  --------------------- (1)

Let   OA = OB = x metres

So            x2 + x2 = 502

Or             2x2 = 50 x  50

X2  = 25 x 50                                        -----------------------(2)

Now ,

Area of sector OAB  = `(90)/(360)` *`Pi` *x2

=`(1)/(4)` * `Pi` i*x2

= `(1)/(4)`  x `(22)/(7)` x 25 x 50m2     [from (2)] ------(3)

Also, area of  `Delta` OAD = `(1)/(4)` * 50 * 50 m2  (<AOB=90)------(4)

So, area of flower bed                 AB = (`(1)/(4)`* `(22)/(7)`*25*50 – `(1)/(4)`*50*50)m2   [from (3) and (4)]


= `(1)/(4)`*25*50(`(22)/(7)` -2)m2

=`(1)/(4)`*25*50* `(8)/(7)` m2   ----------------(5)

Similarly area of the other flower bed

= `(1)/(4)` * 25 * 50 * `(8)/(7)` m2    -------------------------(6)

Therefore,

Total area  =(50*50 + `(1)/(4)` *25*50* `(8)/(7)` +`(1)/(4)`*25*50*`(8)/(7)`)m2    [from (1),(5) and (6)]

=25*50(2+`(2)/(7)`+`(2)/(7)`)m2

=25*50* `(18)/(7)` m2

=  3214. 29 m2

Areas of Combinations of Plane Figure Example 2:

Example 2:

Find the area of the shaded region in the following figure , where ABCD is a square of a side  10 cm.





Solution:

Area of square ABCD

= 10 * 10 cm2

= 100 cm2

Diameter of each circle    = `(10)/(2)`cm=5cm

So, radius of each circle      =(5)/(2) cm

<br>

=`(22)/(7)` *`(5)/(2)` * `(5)/(2)`cm2

=`(550)/(28)` cm2

Therefore area of the four circles = 4* `(550)/(28)`cm2 =  78.57Cm2

Hence area of the shaded region = (100-78.57)cm2= 21.43Cm2

Polynomials Chart

In mathematics, a polynomial is an expression of finite length constructed from variables (also known as indeterminate) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents. For example, x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number. (Source Wikipedia)

In this article polynomial chart we see about basic concepts of polynomial, its types of polynomial ,some example problems

Polynomial Types:

Basic concepts of polynomials:
Polynomial is nothing but algebraic expression and also concept of algebras. More types of polynomials are available in the algebra depends on the number of terms. Based on the number of terms polynomial was classified four types

Different types of polynomial:

Polynomial chart



Monomial:

If the expression having one term mean it was called as monomial

Example: 7x ,8x2

Binomial:

If the expression having two terms mean it was called as binomial

Example: 6x+4x

Trinomial

If the expression having three terms mean it was called as trinomial

Example: 3x+8x2+9

Polynomial:

If the expression having more than three terms mean it was called as polynomial

Example: 5x2+12x3+9x+10

Polynomial operations are addition of polynomial, subtraction of polynomial, multiplication of polynomial, division of polynomial.

Example Problems in Polynomial:

Example problems in Polynomial degree chart:

Polynomial addition chart:

Example 1:

Add the polynomial 3x2+5x+2 and 5x+6

Given polynomials: 3x2+5x+2,5x+6

Now we have to arrange the  terms for addition

After than add the terms one by one.

This is a polynomial addition chart



Example 2:

Polynomial multiplication chart:

(2x+5)(3x+1)

Now we have to multiply the one terms with another terms

And then add the terms



Example 3:

Degree of polynomial:

(9z9 +8 z4 − 6z5 + 8) Find the degree of polynomial for each term?

Degree of polynomial for first term=9

Degree of polynomial for second term =4

Degree of polynomial for third term=5

Degree of polynomial for fourth term=0

Highest degree of polynomial is 9

Ratio to Fraction Converter

Ratio :

In mathematics, The ratio can be used to relate two quantities by using the symbol : Also it can be expressed as follows,

• x is to y
• the ratio of x to y
• x : y

Fraction :

In mathematics , Part of the whole can be expressed as fraction. There are three kinds of fraction

• Proper fraction
• Improper fraction
• Mixed fraction

In this article we are going to see about how to simplify the ration as fraction by using the ratio to fraction converter.

Ratio to Fraction Converter :

Converter:

The electronic or software device that can perform the operations Quickly. The ratio to fraction converter can be used to convert fraction for the given ratio.



Fig(i) Ratio to fraction converter

Let us see some problems on ratio to fraction convertor.

Problems on Ratio to Fraction Converter :

Problem 1:

Convert the ratio 45 : 180 into simplified fraction

Solution:

Given,The ratio 45 : 180

We need to convert the given ratio into fraction .

we know that 45 : 180 = ` 45/ 180`

Divided by  45 on both numerator and denominator,

`45/180` = `( 45 / 45 ) / ( 180 / 180 )`

= `1 / 4`

Answer: The simplified fraction of the given fraction is  `1/4` .


Problem 2:

In a bag, there is Blue  and Green balls, the ratio of Blue balls to Green balls is 5:6. If the bag contains 180 Blue balls, how many green balls are there?

Solution:

Given The ratio of the Blue and green balls = 5 : 6

Number of blue balls = 180

Let us take x = green balls

To find the green balls we need to convert the given ratio into fraction,

Write the items in the ratio as a fraction.

`(blue) / (green)` = `5/6` = `x / 180`

`5/6` = `x / 180`

Multiply by 6 on both sides,

5 = `x / 180 `

5 = `x / 30`

Now multiply by 30 on both sides,

5 * 30 = x

150 = x

x = 150

Total number of black balls = 150

Answer: Green balls = 150