Quadratic Inequalities

In this page we are going to discuss about  quadratic inequalities concept.A quadratic inequality in one variable is in the form of an expression:

ax2 + bx + c ≤ 0  or   ax2 + bx + c < 0
ax2 + bx + c ≥ 0  or   ax2 + bx + c < 0

where a, b and c are real numbers, a ≠ 0. The values of x which satisfy the given inequality are called the solutions of the inequality.

A quadratic equation has only two roots. But a quadratic inequality has many roots.


Methods to solve quadratic Inequalities


There are two methods to solve quadratic inequalities -

Method 1: Finding the solution by dividing the given polynomial into factors. This method is called the 'Algebraic method'.

Method 2: Finding  the solutions by drawing the graph of the inequality. This method is called the 'Graphical method'.

Note:

• The trick in solving a quadratic inequality is to replace the inequality symbol with an equal sign and solve the resulting equation.  The solutions to the equation will allow us to establish intervals that will let you solve the inequality.
• Plot the solutions on number line creating the intervals for investigation. Pick any number from each interval and test it in original inequality. If the result is true, that interval is the solution to the inequality.

Solving quadratic inequalities


Below you can see the example on solving quadratic inequalities-

Example:1 Find the solution set of x2 – 4x – 21 ≥ 0 in

1. Algebraic Method
2. Graphical Method

Solution:

1.  Algebraic Method:

X2 - 4x - 21 ≥ 0
=> ( x + 3) ( x - 7 ) ≥ 0
Here the coefficient of x2 is > 0 and the sign of the quadratic expression is non-negative.
So, x ≥ -3 and x ≤ 7 or there are two cases for the product ( x + 3 ) ( x - 7 ) to be non-negative.

Case (1):
x + 3 ≥ 0 and x - 7 ≥ 0
=> x ≥ -3 and x ≥ 7
So it is necessary that  x≥ 7. Notice the line marked below with x ≥ -3 and x ≥ 7 and their intersection.



From the figure above, it is clear that the intersection is x ≥ 7.

Case (2):
x + 3 ≤ 0 and x - 7 ≤ 0
=>   x ≤ -3 and x ≤ 7
So it is clear x ≤ -3.
i.e., the values of x for which x2 - 4x - 21 ≥ 0 are given by these two cases.

All values of x satisfying the in equations x ≤ -3 or x ≥ 7 become the solution set for the given in equations. This solution set can  be shown in the graph given below.



From the figure above, it is clear that the intersection is x ≥ 7.


2. Graphical Method:

The given polynomial function is f(x) = x2 - 4x - 21

x	-3	-2	-1	0	1	2	3	4	…
f(x)	0	-9	-16	-21	-24	-25	-24	-21	…

Using the above chart, we can draw a graph given below





Now, we require the values of x satisfying x2 - 4x - 21 ≥ 0 , i.e., f(x) ≥ 0 ( y ≥ 0 ). That is we require the values of x for which f(x) lies on x-axis and above. But it is clearly understood from the figure that the values of  x are -3, left side to -3, +7, and right side to +7. Thus all values of x such that x ≤ 3 and x ≥ 7 become the solution set.

Review Geometry Learning

Geometry is learning a part of mathematics which deals with the dimensions of solids and surfaces, shapes, angles and lines etc., It reviews the deduction concepts and consequence logic's, which can be applied through out your life time . In this article, you can review the basic terms and shapes in geometry.

Learning Basics Terms in Geometry:

Learning the following definitions helps to review concepts in geometry.

Point:
A point have no dimensions, it just denotes a position.

Line:
A line can be defined as a straight curve which connects many points. A line has only one dimension, i.e., length.

Collinear points:
If two or more points lie on same line, then they are said to be as colinear points.

Plane:
A Plane can be defined as a set of points and line segments combined infinitely to form a flat surface, it can be extended infinitely in any directions. A plane has infinite length and width, but it doesn't have any height.

Midpoint:
A midpoint can be defined as a point, which divides a line segment equally from both the end points.

Line segment:
A line segment can be defined as a part of a line. It consists of two end points and it was represented by its end points.

Ray:
A ray can be defined as a line having one fixed end point and an infinite extension on the other end.

Angles:

An Angle can be formed by two rays having a common end point.These are the reviews of the basic terms in geometry learning.

Learning Basic Shapes in Geometry:


Square:
A Square can be defined as a regular quadrilateral with equal sides and 4 right angles.

Rectangle:
A Rectangle is a four sided geometrical figure with two pairs of same side lengths.

Triangle:
In geometry, only shape which has three sides and angles is a triangle.

Circle:
A circle is a geometric figure, formed by a locus of points which are all equidistant from a common point called as its center. The distance between the locus of points to the common point is known as radius of the circle.

These are the reviews of the basic shapes in geometry learning.

Long Integer Definition

Integer is a whole number which is not a fraction, that is it can be either positive, negative or zero. Thus the numbers -20, -10, 0, 10, 50, 98, 540 are integers. It can not have decimals. It is commonly used in computer programming as a data type. It can also be used to find out item's location in an array. If two integers are added, subtracted or multiplied, then the output is also an integer. Where as if one integer is divided with other integer, then the result may be an integer or a fraction.

When we think of the number, we first of all think of natural numbers: 1, 2, 3, 4, 5, ...... which are also called positive integers. By applying the operation of subtraction, we get the number zero and negative integers. The set {.....-4, -3, -2, -1, 0, 1, 2, 3, 4,........} is called the set of integers. By division operation, we get the positive and negative fractions. The integers and fractions together constitute the class of rational number p/q, where p and q are integers but q is not equal to 0.

Integers and long integers:


Numbers which includes positive integers{ 1, 2, 3, 4, .....} and negative integers{-1, -2, -3, -4,......} with zero{0} are called integers. There is no fractions or decimal points are included in the integers. They are represented by Z.

Therefore,

Z={ … -4, – 3, – 2, – 1, 0, + 1, + 2, + 3, +4 ….}

+ 1, + 2, + 3, … are positive integers.

– 1, – 2, – 3, … are negative integers.

Note:

Positive numbers can be written even without the ‘+’ sign.

For example, +5, +4, +3, +2, +1 are  written as 5, 4, 3, 2, 1.

Thus, Z = {…… – 3, – 2, – 1, 0, 1, 2, 3, ……}

The integers are represented on the number line as follows:



On the number line, the numbers which are to the right of zero are called positive integers. The numbers which are to the left of zero are called negative integers.

(1)  Every natural number or a whole number.

(2)  Every whole number is an integer.

Order in long integer:

• On the number line, we find that the number value increases as we move to the right and decreases as we move to the left.
• If we represent two integers on the number line, the integer on the right is greater than the integer on the left.
• In other words, the integer on the left is lesser than the integer on the right.

For example, consider the following points marked on the number line in the figure given below:


In the above figure,

4 is to the right of 2                `:.` 4 > 2

2 is to the right of -1               `:.` 2 > -1

-5 is to the left of  -1                 `:.` -5 < -1


Example problems:


Example 1: Which is smaller? -2 and -5

Solution:

First, mark the integers -2 and -5 on the number line.



On the number line, -5 is on the left side of -2.

Therefore, -5 is smaller than -2

That is -5 < -2.

Example 2: Write the following integers in ascending order.

3,-2,0,-4,-1,5

Solution:

First mark these integers on the number line.



Now arrange the integers from left to right to get them in ascending order

-4 < -2 < -1 < 0 < 3 < 5

Therefore, the ascending order is -4, -2, -1, 0, 3, 5.

Which Fraction is Larger

 Yes, which fraction is larger or smaller? The question has great importance. As in many of the competitive exams, we are asked to arrange a set of given fractions. Before learning, how to compare the fractions, we'll learn the basics of fractions. Then we'll learn two methods, that will help us to tell "which fraction is larger".

Fraction: A number of the form `a/b` , where a and b are integers and b`!=` 0,  is known as fraction. Here, a is called numerator and b is called denominator.

Example: `3/5` is a fraction, where Numerator = 3, and Denominator = 5.

7 is also a fraction, because it can be written as `7/1`

1st Method to find which fraction is larger


Method to compare two fractions: This method is known as Cross multiplication method. Steps are as follows:

Let `a/b` and  `c/d` be the two given fractions, that we have to compare and decide which fraction is larger?

Find the product of numerator of the first fraction and the denominator of the 2nd fraction. So, we get ad and bc.

i) If ad > bc, then `a/b`  >  `c/d` .

ii) If ad = bc, then   `a/b` = `c/d` .

iii) If ad < bc, then `a/b`  <  `c/d`

Ex: Compare, which fraction is larger `3/5` and `5/8`?

Products are 3X8= 24 and 5X5=25.

Here, 24 < 25,  So `3/5`` < ``5/8`

Now, we can say very easily, which fraction is larger.

Hence, `5/8` is larger fraction among the two.


2nd Method to know which fraction is larger:


This method is used to compare more than two fractions:

Step I: Find the LCM or LCD of the given fractions. Let it be m.

Step II: Convert all the given fractions into like fractions, each having m as denominator.

Step III: Now we compare the numerators. The fraction having greater numerator is greater and vice- versa.

Ex: Arrange the fractions  `2/5` , `3/10` and `9/14` in ascending order.

LCM of 5, 10 and 14 is 70.

Now, let us change each of the given fractions into an equivalent fraction having 70 as denominator.




Practice problems of which fraction is larger


1. Which fraction is larger: `5/6` or `6/7` ?

2. Which fraction is larger: `21/25` or `3/5` ?

Study graph functions

Graphs of functions f is the collection of the all ordered pairs (x, f(x)). In particular, if x is a real number, a graph means the graphical representation of this collection, in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. Graphing on the Cartesian plane is sometimes are called curve sketching. If the function input x is an ordered pair (x₁, x₂) of real numbers, the graph is the collection of all ordered triples (x₁, x₂, f(x₁, x₂)), and its graphical representation is a surface .

Example problems for study graph functions :

Study linear functions graph:
If a function f   : R → R is defined in the form f(x) = dx + e then the function is called a linear function. Here d and e are constants.

Problem 1: 

Draw the graph of the linear function f : R → R defined by f(x) = 7x + 1.
Solution:
Draw the table of some pairs (x, f(x)) which satisfy f(x) = 7x + 1.

x	-2	-1	0	1	2
f(x)	15	8	1	-6	-13

Plot the points and draw the curve passing through these points. Note that, the curve is a straight line.



Study functions of graph:

Graph of a functions: The graph of a function f is a graph of the equation y = f(x)

Problem 2:
Draw the graph of the function f(x) =2x²

Solution:
Draw a table of some pairs (x, y) which satisfy y = 2x²

x	-3	-2	-1	0	-1	-2	-3
f(x)	18	8	2	0	2	8	18

Plot the points and draw the smooth curve passing through the plotted points.



Note:

Note that if we draw a vertical line to the above graph, it meets the curve at only one point
i.e. for every x there is a unique y

Example problem for study logarithmic graph functions :

Draw the graphs of the logarithmic functions (1) f(x) = log ₅2x (2) f(x) = log ₁₀ex (3) f(x) = log ₅3x
Sol:

The logarithmic function is the defined only for the positive real numbers. i.e. (0, ∞)
Domain: (0, ∞) Range: (− ∞, ∞)

Study Online Exponentiation

Exponentiation should be the operation, which is written as the form of an. Where a and n is said to be base and exponent as well as n is any positive integer. In general, exponentiation means that repetitive multiplication. Otherwise, exponentiation an is the product of n factors of a. Online study should be the topic-oriented or else technical help for clarifying doubts that are convey through the computer software. Let us study properties and example problems for exponentiation.


Properties - Study Online Exponentiation

We are having seven number of exponentiation properties that used for solving problems. In this properties, a, m and n are any integer values.

Product of like bases:

The product of powers with the same base means we can add the powers and keep the common base.

am an = am+n

Quotient of like bases:

To divide the powers with similar base, we can subtract the exponents and remain the common base.

`a^m/a^n` = am-n

Power to a power:

In case of raising the power to power, we need to keep same base and multiply the exponent values.

`(a^m)^(n)` = amn

Product to a power:

In case of raising the product to power, we need to raise each factor to the power value.

(ab)m = am bm

Quotient to a power:

In case of raising the quotient to power, we need to raise the numerator and denominator to the power value.

`(a/b)^n` = `a^n/b^n`

Zero exponent:

Any number that is raised with zero power should be equivalent to ‘1’.

a0 = 1

Negative exponent:

a-n = `1/a^n` or `1/a^(-n)` = an

These are the properties that are used for exponentiation problems in study math online.


Example Problems - Study Online Exponentiation


Example 1:

Solve 32 34.

Solution:

Given, 32 34.

This is in the structure of am an, so we need to use am an = am+n property.

Here, m = 2 and n = 4 and a = 3.

Thus, 32 34 = 32+4

= 36

= 3 × 3 × 3 × 3 × 3 × 3

= 9 × 9 × 9

= 729

Hence, the answer is 32 34 = 729.

Example 2:

Shorten the following `6^7/6^4` .

Solution:

Given, `6^7/6^4` .

This is in the structure of `a^m/a^n` , so we need to use `a^m/a^n` = am-n property.

Here, m = 7 and n = 4 and a = 6.

Thus, `6^7/6^4` = 67-3

= 64

= 6 × 6 × 6 × 6

= 36 × 36

= 1296

Hence, the answer is `6^7/6^4` = 1296.

That’s all about the study online exponentiation.

Simple Algebra Problems

Algebra is a branch of mathematics dealing with variables, equations , expressions.Here we are going to solve some of the simple algebra problems including some equations, expressions and algebraic identities.

Solve Simple algebra problems on numbers:

Ex 1: Subtract  – 4 from – 10.

Sol :
=– 10 – (– 4)

= – 10 + (additive inverse of – 4)

= – 10 + 4

= – 6 (see addition of two integers)

Ex 2: Find the product of  (a) (+ 6) × (– 5) (b) (– 12) × (+ 12) (c) (– 15) × (– 4)

Sol : 

(+ 6) × (– 5) = – 30 (positive × negative= negative)

(– 12) × (+ 12) = – 144 (negative × positive = negative) 

(– 15) × (– 4)= + 60 (negative × negative = positive) 

Ex 3: Find the sum of given algebraic expression 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Sol:

Using the associative and distributive property of the real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1)

= 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.

  The following scheme is helpful in adding two polynomials 2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1 2x4 + 6x3– 9x2 + 9x + 2 


Ex 4: Find the subtraction of given simple algebraic expression 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Sol: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1)

= x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the sign of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1.

Adding the polynomial in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5 

Ex 5: Find the product of   given simple algebraic expression x3 – 2x2 – 4 and 2x2 + 3x – 1.

Sol:

(x3 – 2x2 – 4) (2x2 + 3x – 1)

= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)

= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)

= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4

= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4

= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.


Algebra Identities to solve some simple problems :


 Algebraic identity for (x + a)(x + b)

 By using the distributive properties of numbers,

(x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab

= x2 + ax + bx + ab= x2 + (a + b)x + ab.

 (x + a)(x + b) ≡ x2 + (a + b)x + ab

(x – a)(x + b) ≡ x2 + (b – a)x – ab

(x + a)(x – b) ≡ x2 + (a – b)x – ab

(x – a)(x – b) ≡ x2 – (a + b)x + ab

(a + b)2 ≡ a2 + 2ab + b2

(a – b)2 ≡ a2 – 2ab + b2

(a + b)(a – b) ≡ a2 – b2

(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3

(a − b)3 ≡a3−3a2b+ 3ab2−b3

a3 + b3 ≡ (a + b)3−3ab(a + b)

a3 − b3 ≡ (a −b)3+ 3ab(a−b)

Solve the following simple algebra problems using above identities:


Pro 1: Find the product:  (x + 3) (x + 5)

Sol: (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15.


Pro 2:  Find the product:  (p + 9) (p – 2)

Sol: (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2

= p2 + 7p – 18.


Solve simple algebra problems on the cubic identity


Algebra problems on cubic identity:

(i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3

= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3

= 27x3 + 54x2y + 36xy2 + 8y3.

 (ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3

= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3

= 8x6– 36x4y + 54x2y2 – 27y3


Pro 1: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.

Solution:a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4)

= 64 – 12

= 52.


Pro 2: If a – b = 4 and ab = 2, find a3 – b3.

Solution:a3 – b3 = (a–b)3+3ab(a–b)

= (4)3+3(2)(4) =64+24

=88.