Quadratic Equation x Intercept

In coordinate geometry, the x intercept means the x-value of the point where the graph of a function or relation intercepts the x-axis of the coordinate system. It means the point at which the line cuts the x-axis.

The x intercept of a line is denoted as (x, 0).

In this article, we are going to learn finding x intercept for the quadratic equation.


Example Problems to Find X Intercept for Quadratic Equation:

Example 1:

Find x intercept for the quadratic equation y = x2 + x - 12

Solution:

Step 1: Given function

y = x2 + x - 12

Step 2: Plug y = 0 in the given function

0 = x2 + x - 12

We can write it as

x2 + x - 12 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for x and - 12x2

x = 4x - 3x

-12x2 = (4x)(-3x)

Replace the term x by 4x -3x

x2 + 4x - 3x - 12 = 0

Take the common factor outside,

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

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

Factor each term,

x = - 4   and x = 3

Step 4: Solution

X intercepts for the given equation are (- 4, 0) and (3, 0)

Example 2:

Find x intercept for the quadratic equation y = 2x2 + 10x + 8

Solution:

Step 1: Given function

y = 2x2 + 10x + 8

Step 2: Plug y = 0 in the given function

0 = 2x2 + 10x + 8

We can write it as

2x2 + 10x + 8 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for 10x and 16x2

10x = 8x + 2x

16x2 = (8x)(2x)

Replace the term 10x by 8x + 2x

2x2 + 8x + 2x + 8 = 0

Take the common factor outside,

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

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

Factor each term,

x = - 1   and x = - 4

Step 4: Solution

X intercepts for the given equation are (- 1, 0) and (- 4, 0)




Practice Problems to Find X Intercept for Quadratic Equation:

1) Find x intercepts for the quadratic equation y = x2 + 18x + 32

2) Find x intercepts for the quadratic equation y =2 x2 + 14x + 12

2) Find x intercepts for the quadratic equation y = 3x2 + 6x - 8

Solutions:

1) x intercepts are (-2, 0) and (-16, 0)

2) x intercepts are (-1, 0) and (-6, 0)

3) x intercepts are (0.914, 0) and (-2.914, 0)

Histogram Unequal Intervals

The Histogram Definition unequal intervals of applications are mostly used in statistics. The histogram unequal intervals are most effective way  to analysis the frequency distribution of grouped data. The concepts of histogram unequal intervals to show the quick way of understanding the grouped data in the graphical representations. The analyzing histograms diagrams are constructed from the two ways are depending on the uniform width, and varying widths.
Constructions for Histogram Unequal Intervals:

To analyzing  the histogram unequal intervals , we draw two perpendicular axes and choose a suitable scale for each axis. We blot class or group intervals of the continuous grouped or class of data on the horizontal axis and the respective number of  occurrence  of each class on the vertical axis. For each class, a rectangle is constructed with class interval as the base and height determined from the class frequency so that areas of the rectangles are proportional to the frequencies.

Let us analyzing histogram unequal intervals through an example, how a grouped frequency distribution of number of teachers in a city can be represented in histogram.


Histogram

From the above histogram unequal intervals diagrams, it is clear that the maximum number of students in the group 50 - 60 and the minimum number of students in the group 70 - 90.

Notes: In the diagram  (kink) before the class interval 30 - 40 on the horizontal axis. It shows that the full distance 0 - 30 is not shown.
Example of Histogram Unequal Intervals:

Another example of histogram unequal intervals for a math tutor wanted to analyse the performance of two class of students in a math test of 100 marks. Looking at their performances,  math tutor found that a few students got under 20 marks and a few got 70 marks or above. So the math tutor decided to analyzing histogram unequal intervals to group them into intervals of varying sizes as follows: 0 - 20, 20-30, ...., 60 - 70, 70 - 100. Then she formed the following table.
Marks                          Number of students
0 - 20                                         9
20 - 30                                       14
30 - 40                                       14
40 - 50                                       18
50 - 60                                       18
60 - 70                                       12
70 - above                                  4

Negative Fraction Calculator

Negative fraction is nothing but, the given input fractions are in negative values, and using negative fraction calculator will produce the output with in a second. Numerator and denominator to created the fraction, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole.


Example is 4/5, the numerator value, 4, tells us that the fraction represents 4 equal parts, and the denominator, 5, tells us that 5 parts make up a total. An afterward growth be the common or "vulgar" fractions which are still used today like this (½, ⅝, ¾, etc.).
Examples Problems for Using Negative Fraction Calculator:

1. Subtract the fraction of -5/6 - 8/6, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click the enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -5/5 - 2/6

= (-5 - 2) / 6

We get the answer is,

= -7 / 6.

2. Subtract the fraction of -9/2 - 8/2, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -9/2 - 8/2

= (-9 - 8) / 2

We get the answer is,

= -17 / 2.

More Examples Problems for Using Negative Fraction Calculator:

3. Subtract the fraction of -2/7 - 4/7, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -2/7 - 4/7

= (-2- 4) / 7

We get the answer is,

= -7 / 6.

4. Subtract the fraction of -6/3 - 5/3, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,



fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -6/3 - 5/3

= (-6 - 5) / 3

We get the answer is,

= -11 / 3.

Linear Functions and Matrices

Definition of Matrices:

A matrix is a group of numbers arranged into a fixed quantity of rows and columns. Usually a matrix is a two dimensional range of numbers or terms arranged in a set of rows and columns. m × n matrices A have m rows and n columns and it is written,

A = `[[a_(11),a_(12),a_(1n)],[a_(21),a_(22),a_(2n)],[a_(m1),a_(m2),a_(mn)]]`

Definition of Linear Functions:

A polynomial functions of single degree is defined as a linear functions. It relates a dependent variable with an independent variable in a simple way. Mathematical equation in which there is no independent-variable is raised to a power greater than one. A simple linear function with one independent variable traces a straight line when plotted on a graph. It is also called as linear equation.
Solving Linear Functions Using Matrices:

Steps to solve the system of linear equations using matrices,

Step 1: Let us consider the equation,

a1x + b1y = c1,

a2x + b2y = c2,

Step 2: If, A = `[[a_1,b_1],[a_2,b_2]]` , X = `[[x],[y]]` , and C = `[[c_1],[c_2]]` .

then, AX = C.

Step 3: Now multiply both sides by, A−1

=> A−1AX = A−1C

Step 4: We know that, A−1A = I, where I is identity matrices,

=> IX = A−1C.

Step 5: Now, Multiplication of any matrices with identity matrices gives the same matrices. So,

Hence, X = A−1C.
Example for Solving Linear Functions Using Matrices:

Example: 5x – y + 11 = 0 and x – y – 5 = 0, solve the linear equations for x and y using matrices.

Solution:           5x – y + 11 = 0     => 5x – y = – 11,

x – y – 5 = 0         => x – y = 5.

Let, A =` [[5,-1],[1,-1]]` , X = `[[x],[y]]` , and C = `[[-11],[5]]`

To find the value of A−1:

A = ` [[5,-1],[1,-1]]`

Swap the leading diagonal values,

=> ` [[-1,-1],[1,5]]`

Change the signs of the other two elements in the matrices,

=>` [[-1,1],[-1,5]]`

Now, find the value of |A|.

|A| = (–1 × 5) – ((–1) × 1)

=> |A| = –5 + 1

=> |A| = –4

Hence, A−1 = `1/(|A|) xx [[-1,1],[-1,5]]`

=> A−1 = `1/(-4) xx [[-1,1],[-1,5]]`

=> A−1 =` [[(-1)/(-4),1/(-4)],[(-1)/(-4),5/(-4)]]`

=> A−1 =` [[1/4,(-1)/4],[1/4,(-5)/4]]`

Therefore the solution is,

X = A–1C,

=> X =` [[1/4,(-1)/4],[1/4,(-5)/4]] xx [[-11],[5]]`

=> X =`[[(-5)/(15) -(10)/(15)],[(10)/5 - 5/5]]`

=> X = `[[(-15)/(15)],[5/5]]`

=> X =` [[-1],[1]] = [[x],[y]]`

Hence the solutions are, x = – 1, y = 1.

Solve Herons Formula

We are familiar with the general formula for finding the area of a triangle when the base and height of the triangle is given. For scalene traingle we do not have any area formula. Heron's formula proof is a general formuls used to find area of triangles of all types.

Heron a famous mathematician gave simple formula for finding the area of any triangle based on its three sides. Thus, this formula for area is known as Heron's formula and is stated as below,

If p, q, r, denote the lengths of sides of a `Delta` PQR, then,

Heron's formula of a triangle

Area of `Delta` PQR = `sqrt(s(s-p)(s-q)(s-r))` , where   s = `(p+q+r)/(2) = ("perimeter of triangle")/(2)`

Here, s = semiperimeter of `Delta` PQR.

Heron's formula can used to find area of all types of triangles, quadrilateral, trapezoid, etc.
Solved Examples Using Heron's Formula

Ex 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm, and 200 cm.

Sol: Given three sides, p = 150 cm, q = 120 cm and r = 200 cm.

Step 1:  s = `(p + q + r)/(2) = (150 + 120 + 200)/(2)`

= `(470)/(2) = 235`

Step 2:  (s-p) = (235 -150) = 85

(s-q) = (235 - 120) = 115

(s -r ) = (235 - 200) = 35

Step 3: Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt(235xx 85xx 115xx 35)`

= `sqrt(5xx 47 xx 5xx 17 xx 5 xx 23 xx 5 xx 7)`

=  25`sqrt(47 xx 17 xx 23 xx 7)`

= 8966.56 cm2

Ex 2: Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42 cm.

Sol:  Given p = 18 cm, q = 10 cm and perimeter = 42 cm

Step 1:  perimeter = p + q + r

42 = 18 + 10 + r

42 = 28 + r

subtract 28 on both sides

42 - 28 = 28 - 28 + r

14 = r

Step 2 :  s = `("perimeter)/(2) = (42)/(2)`

s = 21

Step 3:  (s - p) = 21 - 18 = 3

(s - q) = 21 - 10 = 11

(s - r)  = 21 - 14 = 7

Step 4:  Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt ( 21 xx 3 xx 11 xx 7)`

= `sqrt( 3 xx 7 xx 3 xx 7 xx 11)`

= 3 x 7 `sqrt(11)`

= 21`sqrt(11)` cm2
Practice Problems on Heron's Formula

Pro 1: Find the area of a triangle whose sides are respectively 100 cm, 80 cm, and 60 cm

Ans: 2400 cm2

Pro 2: Find the area of a triangle whose two sides are 12 cm and 8 cm and the perimeter is 30 cm.

Ans: 39.68 cm2

Adding Cosine Functions

Trigonometry is one of the part of mathematics managing with angles, triangles and trigonometry functions such as sine, cosine, and tangent which are abbreviated as sin, cos, and tan respectively. There are several formulas for adding trigonometric functions such as sine, cosine, and tangents. The formulas for adding cosine functions are used to find the ratios when two angles are given. Those formulas for adding cosine functions are shown below.


cosine (A + B) = cosine A cosine B − sine A sine B.
Proof for Adding Cosine Functions:

To prove: cosine (A + B) = cosine A cosine B − sine A sine B.

Proof: We know that, cosine (A − B) = cosine A cosine B + sine A sine B.

Replacing B with (−B), these identities becomes,

sine (A − (− B)) = cosine A cosine (− B) + sine A sine (− B)

Since we know that, cosine (−A) = cosine A, and

sine (−A) = − sine A.

cosine (A + B) = cosine A cosine B − sine A sine B.

Hence proved that, sine (A – B) = sine A cosine B – cosine A sine B.
Solved Examples for Adding Cosine Functions:

Example 1: If sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8, then find cosine(A + B).

Solution: Given that, sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (12 × 8) − (14 × 6),

=> cosine (A + B) = 96 − 84,

=> cosine (A + B) = 12.

Answer: cosine (A + B) = 12.

Example 2: If sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3,  then find cosine(A + B).

Solution: Given that, sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (17 × 3) − (5 × 9),

=> cosine (A + B) = 51 − 45,

=> cosine (A + B) = 6.

Answer: cosine (A + B) = 6.

Rules for Division

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.

Understanding Rules of Divisibility is always challenging for me but thanks to all math help websites to help me out

Specifically, if c times b equals a, written:

C = b x a

where b is not zero, then a divided by b equals c, written:

a/b = c

For instance,

6/3 = 2

since

2 x 3 = 6

In the above expression, a is called the dividend, b the divisor and c the quotient.
Basic Rules for Division:

The basic rules for division are given as below:

Rule 1:

Division of 2 numbers with the same sign should be ‘+’ ve (positive sign) sign.

‘+’ve (Positive number) ÷’+’ve ( positive number) = ‘+’ve ( positive number)
‘-‘ve (Negative number) ÷ ‘-‘ ve (negative number) = =’+’ ve (positive number)

Rule 2:

Division of 2 numbers with different signs should be ‘-’ve negative

‘+’ve (Positive number) ÷(‘-‘ve negative number) = ‘-‘ve (negative number)
‘-‘ve (Negative number) ÷ ‘+’ve (positive number) = ‘-‘ve (negative number)

Examples for division:

Examples:

54 ÷ 9 = 6 (same signs)

(-32) ÷ (-2) = 16 (same signs)

10 ÷ (-2) = -5 (different signs)

(-12) ÷ 12 = -1 (different signs)


Rules for Division on Dividing Variable:

In this operation division represents dividing the variables based on the presence of values.

Example:

Solve the following division: 49 ÷ p, given that p = -7

Solution:

49 ÷ p

Substitute p = -7,

= 49 ÷ (-7)

= -7 (dissimilar signs).
Division on Dividing Decimal on Rules for Division:

Rules for decimal division solving problems,

Rule 1:

To create the decimal divisor as whole number by changing the decimal point to the right side.

Rule 2:

To change the same decimal point in the dividend to the right side to create as whole number

Rule 3:

After divide the new dividend or whole number by new divisor or whole number


Example:

Divide the following decimal function:  24.24 ÷ 0.20

Solution:

24.24 ÷ 0.20 = 24.24 / 0.20

= 242.4 / 2 (Take the decimal divisor as whole number)

= 2424 / 20 (Take the decimal dividend as whole number)

= 121.2 (Divide the new dividend by new divisor).