Square root and Approximations

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Square root and Approximations

To know about approximating square roots, first we should learn perfect squares because they are interrelated. Here is a table of perfect squares and square roots.

 \dpi{120} 1^{2}=1 \dpi{120} \sqrt{1} =1
\dpi{120} 2^{2}=4 \dpi{120} \sqrt{4} =2
 \dpi{120} 3^{2}=9 \dpi{120} \sqrt{9} =3
\dpi{120} 4^{2}=16  \dpi{120} \sqrt{16} =4
 \dpi{120} 5^{2}= 25  \dpi{120} \sqrt{25} =5
\dpi{120} 6^{2}= 36  \dpi{120} \sqrt{36} =6
\dpi{120} 7^{2}= 49  \dpi{120} \sqrt{49} =7
\dpi{120} 8^{2}= 64  \dpi{120} \sqrt{64} =8
\dpi{120} 9^{2}= 81  \dpi{120} \sqrt{81} =9
\dpi{120} 10^{2}=100  \dpi{120} \sqrt{100} =10
\dpi{120} 11^{2}= 121  \dpi{120} \sqrt{121}= 11
\dpi{120} 12^{2}=144  \dpi{120} \sqrt{144}=12
\dpi{120} 13^{2}=169  \dpi{120} \sqrt{169} = 13
\dpi{120} 14^{2}=196  \dpi{120} \sqrt{196}= 14
\dpi{120} 15^{2}= 225  \dpi{120} \sqrt{225}=15
\dpi{120} 16^{2}=256  \dpi{120} \sqrt{256}=16
\dpi{120} 17^{2}=289  \dpi{120} \sqrt{289} = 17
\dpi{120} 18^{2}=324  \dpi{120} \sqrt{324} = 18
\dpi{120} 19^{2}= 361  \dpi{120} \sqrt{361} = 19
\dpi{120} 20^{2}=400  \dpi{120} \sqrt{400} = 20

 

This way  you can learn the square roots upto 50. Lets discuss the process of approximating square roots now.

There are many ways to approximate the square roots. The one which we are going to discuss, is the easiest and quicker one. Here are the steps:

1)First find the nearest perfect squares, in which the sqrt lie between.

2) Lets assume the sqrt number to be approximated as S, smaller perfect square as N and larger perfect square as N+1 such that

\dpi{120} N^{2}< S < (N+1)^{2}

Then the approximation A is given by the following formula.

\dpi{120} \mathbf{A=N+\left ( \frac{S-N^{2}}{2N+1} \right )}

Lets work on some examples to understand the process

 

Example1. Approximate √34   to one decimal place.

Solution:  √34  lie between √25  and √36  which are nearest perfect squares so,

\dpi{120} \sqrt{25}< \sqrt{34}< \sqrt{36}

\dpi{120} 5^{2}< 34 < 6^{2}

Here S= 34 , N=5 ,  N+1=6

Using formula,   \dpi{120} A=N+\left ( \frac{S-N^{2}}{2N+1} \right )

\dpi{120} A=5+\left ( \frac{34-5^{2}}{2*5+1} \right ) = 5 +\frac{9}{11} = 5.8

 

Example2. Approximate  √70  to one decimal place.

Solution:    √70  lie between √64  and  √81   which are the nearest perfect squares so,

\dpi{120} \sqrt{64}< \sqrt{70}< \sqrt{81}

\dpi{120} 8^{2}< 70 < 9^{2}

Here S= 70 , N=8 ,  N+1=9

Using formula,    \dpi{120} A=N+\left ( \frac{S-N^{2}}{2N+1} \right )

\dpi{120} A=8+\left ( \frac{70-8^{2}}{2*8+1} \right ) = 8 +\frac{6}{17} = 8.4

 

The method discussed above gives you sqrt approximation upto one decimal place.

Second method: If you are on time constraint For SAT, ACT or any other test and you are given  choices then we can fairly guess the right answer by finding the nearest perfect squares and distance of  sqrt from the nearest perfect square.

For example we need to approximate  √70  which lie between perfect squares 64 and 81.

\dpi{120} \sqrt{64}< \sqrt{70}< \sqrt{81}

\dpi{120} 8< \sqrt{70}< 9

Distance between 70 and its nearest perfect square 64  is 6 which is less then halfway (72.5) that means √70 would be approximated to a value less than 8.5 but closer to 8.5. Based on this observation we can guess the right answer out of the following options.

a) 7.8

b) 8.6

c) 8.4

d) 8.9

We are pretty sure about the right answer to be 8.4 because Only this option is less than 8.5 and closer to it.

To approximate √74  we choose 8.6 because 74 is more than halfway  72.5  and closer to it.

To approximate √ 79 we choose 8.9 because 79 is more than halfway 72.5 and more closer to 81 so it would be approximated closer to 9.

 

So when we are on a time constraint and have multiple choices , we can use above estimation process.

First method gives you  the exact approximation upto one decimal place , second method also gives a pretty good estimation  within seconds.

You can always check your results by using calculators.

 

 

Operations on Scientific Notations worksheet

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Addition and subtraction in scientific notation:

While performing basic arithmetic operations like addition, subtraction, multiplication or division on   numbers written in scientific notation, we need to be careful and should follow some rules to reach the correct answer.

While adding and subtracting the numbers in scientific notation, exponent part of all the numbers must be same. If it is not same , then first make it same and then combine the numbers.

Don’t forget to write the final answer in scientific notation.

When exponent part  is same-

Example1: Combine \dpi{120} {\color{Red} 4.52\times 10^{3}}   and  \dpi{120} {\color{Red} 8.3\times 10^{3}}

Solution: Here we see that exponent part of both numbers is same (\dpi{120} 10^{3}). So we just need to combine the coefficients of these two numbers.

\dpi{120} 4.52\times 10^{3} + \dpi{120} 8.3\times 10^{3}   =  \dpi{120} (4.52+8.3)\times 10^{3}

= \dpi{120} 12.82\times 10^{3}

As the coefficient part can’t exceed 10, so we need to convert this  result  into scientific notation.

=  \dpi{120} 1.282\times 10^{4}

 

When exponent part is not same-

Example2: Simplify the following expression:   \dpi{120} {\color{Red} 8\times 10^{8}- 5\times 10 ^{6}}

Solution: As the exponent part is not same, we make them same first.

Again you can do it two ways. Either lower the exponent  \dpi{120} 10^{8}  to \dpi{120} 10^{6}   or  you can raise the exponent \dpi{120} 10^{6}   to  \dpi{120} 10^{8}

First way:  raise the exponent \dpi{120} 10^{6}   to  \dpi{120} 10^{8}    

As we know, when decimal is moved to left, exponent is positive (increased). We need to raise the exponent by 2 so decimal is moved towards left by 2 places and we get \dpi{120} 5\times 10^{6} = 0.05\times 10^{2}\times 10 ^{6}

= \dpi{120} 0.05\times 10^{8}

\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}   =  \dpi{120} 8\times 10^{8}- 0.05\times 10 ^{8}

= \dpi{120} (8-0.05)\times 10^{8}

=  \dpi{120} 7.95\times 10^{8}

Second way:  Either lower the exponent  \dpi{120} 10^{8}  to \dpi{120} 10^{6}

When decimal is moved to right, exponent is negative (decreased). We need to lower the exponent by 2 so decimal is moved towards right by 2 places  and we get    \dpi{120} 8\times 10 ^{8} = 8\times 10^{2}\times 10^{6} = 800\times 10^{6}

\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}  = \dpi{120} 800\times 10^{6}- 5\times 10 ^{6}

=  \dpi{120} 795\times 10^{6}

But this result is not in scientific notation. To get it in scientific notation decimal should be placed after first significant digit.

So final answer is   \dpi{120} 7.95\times 10^{8}

 

Example3. Simplify the expression  \dpi{120} {\color{Red} 9\times 10^{5} + 6\times 10 ^{7}}

Solution:   \dpi{120} 9\times 10^{5} + 6\times 10 ^{7}  =   \dpi{120} 9\times 10^{5} + 6\times 10 ^{2}\times 10^{5}

=   \dpi{120} 9\times 10^{5} + 600\times 10^{5}

= \dpi{120} (9+600)\times 10^{5}

=  \dpi{120} 609 \times 10^{5}

Then we convert it into scientific notation and get answer as  \dpi{120} 6.09 \times 10^{7}

 

Example4. Simplify the expression  \dpi{120} {\color{Red} 9\times 10^{5} - 6\times 10 ^{7}}

Solution:   \dpi{120} 9\times 10^{5} - 6\times 10 ^{7}   =   \dpi{120} 9\times 10^{5} - 6\times 10 ^{2}\times 10^{5}

=   \dpi{120} 9\times 10^{5} - 600\times 10^{5}

=  \dpi{120} (9-600)\times 10^{5}

=    \dpi{120} -591 \times 10^{5}

Rewriting it as scientific notation we get the final answer as   \dpi{120} -5.91\times 10^{7}

 

 

Practice problems:

Simplify each problem and write the answer in scientific notation.

  1. \dpi{120} 1\times 10^{6} - 8\times 10 ^{4}
  2. \dpi{120} 7\times 10^{7} + 3\times 10 ^{4}
  3. \dpi{120} 4\times 10^{8} - 9\times 10 ^{7}
  4.  \dpi{120} 3\times 10^{-2} - 6\times 10 ^{-5}

 

Answers:

1) \dpi{120} 9.2\times 10^{5}

2) \dpi{120} 7.003\times 10 ^{7}

3) \dpi{120} 3.1\times 10^{8}

4) \dpi{120} 2.994\times 10^{-2}

Scientific and Standard Notations

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Scientific and standard Notations

When we think about writing very large or very small numbers in an easy and quicker way, scientific notation comes to our rescue.

Therefore, scientific notation is a way to represent very small or very large numbers. It is also easy to compare many such numbers when written in scientific notations

How scientific notation is written:

Scientific notation has two parts, first one is the coefficient and second part consists of base 10 and its exponent.

Coefficient is always a number between 1 and 10 . It may contain one or more than one significant digits with a decimal.

Base is always 10

Exponents is always an integer.

Converting from standard notation to scientific notation

When we move towards left, exponent is positive.

When we move towards right, exponent is negative.

Example1. Convert 450000 to scientific notation.

Solution: Since the coefficient must lie between 1 and 10 so decimal should be placed after 4. As decimal is moved 5 steps towards left we use positive exponent (5)

So the scientific notation is written as

\dpi{120} 4.5\times 10^{5}

 

Example2. Convert 0.000000215 to scientific notation.

Solution: This is very small number less than 1 so exponent would be negative.

Since coefficient must lie between 1 and 10 so we put decimal after first significant digit which is 2. For that we need to move decimal to 7 places right and whenever we move towards right while converting standard to scientific notation, we use negative exponents.

So the scientific notation is written as,

\dpi{120} 2.15\times 10^{-7}

 

Example3. Express 0.8 in scientific notation.

Solution: Since decimal should be placed after first significant digit so decimal is moved to right by only one place.

= \dpi{120} 8.0\times 10^{-1}

Or it can also be written as   \dpi{120} 8\times 10^{-1}

 

Example4. Expression 30 in scientific notation.

Solution: Coefficient is the first significant digit which is 3 here, so we need to move decimal towards left by one place.

= \dpi{120} 3.0\times 10^{1}

So 30 can be expressed in scientific notation as \dpi{120} 3\times 10^{1}

 

Converting from scientific notation to standard form

For given negative exponents we move towards left.

For given positive exponents we move towards right.

 

Example5. Convert \dpi{120} {\color{Red} 4\times 10^{-5}} to standard notation.

Solution: Since decimal is not given here so we assume 4 as 4.0 and because of negative exponent(-5) we move 5 places towards left.

\dpi{120} 4\times 10^{-5}  =

So the standard form of given number is 0.00004

 

Example6. Convert \dpi{120} {\color{Red} 3.27\times 10^{4}} to standard form.

Solution: Since exponents is given positive so we move decimal towards right by 4 places.

\dpi{120} 3.27\times 10^{4} =

So the standard form of given number is 32700.

 

Example7. Express \dpi{120} {\color{Red} 5.2\times 10^{0}} to standard form.

Solution: Since exponent is 0 that means we don’t need to move decimal. Moreover  \dpi{120} 10^{0} =1    so the given number remain same without any change in it.

\dpi{120} 5.2\times 10^{0}= 5.2\times 1= 5.2

 

 

Practice problems:

Express the given numbers in scientific notation.

  1. 850,000
  2. 0.0000917
  3. 5102000

Express the given numbers in standard form.

  1. \dpi{120} 4.28\times 10^{-2}
  2. \dpi{120} 3.162\times 10 ^{4}
  3. \dpi{120} 1.2\times 10 ^{-5}

 

Answers:

  1. \dpi{120} 8.5\times 10^{5}
  2. \dpi{120} 9.17\times 10^{-5}
  3. \dpi{120} 5.102\times 10 ^{6}
  4. 0.0428
  5. 31620
  6. 0.000012

 

Linear inequalities with radicals

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Linear inequalities with radicals

While dealing with inequalities, we also need to consider the domain of radical expressions. We take into consideration both the solutions, one which we get after solving inequality by squaring both sides and other by setting radicand ≥ 0. lets work on some examples to understand this process better.

Example1. Solve the following inequality and write the solution in interval form.

 {\color{Red} \sqrt{x-1}+2\leq 5}

Solution:  \sqrt{x-1}+2\leq 5

\sqrt{x-1}\leq 3

Squaring both sides,   (\sqrt{x-1})^{2}\leq 3^{2}

x – 1  ≤ 9

x ≤ 10 ——(i)

Since radicals are not defined for negative values so radicand must be positive. Therefore, x-1 ≥ 0

=> x ≥ 1 ——–(ii)

Combining the solutions (i) and (ii), we get final solution as,

1 ≤ x ≤ 10 => [1 ,10]

 

Example2. Solve the following inequality and write the solution in interval form.

{\color{Red} 4-\sqrt{x-3} \geq -2}

Solution:  4-\sqrt{x-3}\geq -2                                     Also,          x-3 \geq 0

-\sqrt{x-3}\geq -6                                                                                    x ≥ 3

\sqrt{x-3}\leq 6       (multiplied with -1 and inequality sign changed)

squaring both sides, (\sqrt{x-3})^{2}\leq 6^{2}

x -3  ≤ 36

x ≤ 39

combining two solutions we get, 3 ≤ x ≤ 39

Final answer in interval form [3,39]

 

Example3.Solve the following inequality and write the solution in interval form.

\dpi{120} {\color{Red} \sqrt[2]{x}\leq \sqrt[3]{x}}

Solution: We can rewrite the given inequality as,

\dpi{120} x^{1/2} \leq x^{1/3}

Since LCM of denominators 2 and 3 is 6 so we multiply exponents of both sides with 6 and get,

\dpi{120} \left ( x^{1/2} \right )^{6}\leq \left ( x^{1/3} \right )^{6}             =>      \dpi{120} x^{3}\leq x^{2}

\dpi{120} x^{3}-x^{2}\leq 0

\dpi{120} x^{2}(x-1)\leq 0

As  \dpi{120} x^{2}  can never be negative so x-1 ≤ 0

x≤ 1 —— (i)

Also \dpi{120} x^{1/2} , being an even radical is never defined for negative values which restrict the given solution to x ≥ 0 —— (ii)

So final solution, after combining (i) and (ii) is given as,

0≤x≤1  =>  [0,1]

 

Example4. Solve the inequality  \dpi{120} {\color{Red} \sqrt{x}\geq 2 \sqrt[4]{x}}

Solution: Rewriting the given inequality as,

\dpi{120} x^{1/2} \geq 2 x^{1/4}

Since lcm of denominators 2 and 4 is 4, So raising exponents on both sides by 4 and we get,

\dpi{120} \left ( x^{1/2} \right )^{4}\geq (2 \left ( x^{1/4} \right ))^{4}

\dpi{120} x^{2}\geq 2^{4}x

\dpi{120} x^{2}\geq 16x

\dpi{120} x^{2}-16x\geq 0

x ( x-16) ≥ 0

As we know that even radicals are not defined for negative values which restrict domain to x ≥ 0,

Also, x-16≥0 => x≥ 16

So final solution is given as, [16,∞)

 

 

Practice problems:

Solve the following inequalities and write the answer in interval form.

  1. \dpi{120} \sqrt[4]{x}> \sqrt[3]{x}
  2. 3\sqrt{2x-1}+6\leq 15
  3. \dpi{120} \sqrt{c+4}-3\geq 3
  4. \dpi{120} -3\sqrt{11r+3}> -15

Answers:

  1. (0,1)
  2. [1/2, 5]
  3. [32,∞)
  4. ( -3/11 , 2)

 

Linear inequalities with absolute sign

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Linear inequalities with absolute sign are not as easy as  simple linear inequalities. They can be divided into 4 cases depending on the inequality sign and positive or negative right side. All these cases are described in a table given below.

Case 1 Case 2 Case 3 Case 4
 \dpi{150} \mathbf{\left | x \right |}< \mathbf{a}      or

\dpi{150} \mathbf{\left | x \right |}\leq \mathbf{a}

    \dpi{150} \mathbf{\left | x \right |}> \mathbf{a}    or

\dpi{150} \mathbf{\left | x \right |}\geq \mathbf{a}

 

           \dpi{150} \mathbf{\left | x \right |}\leq \mathbf{-a}     \dpi{150} \mathbf{\left | x \right |} > \mathbf{-a}
\dpi{120} -a < x < a         or

\dpi{120} -a\leq x\leq a

After rewriting like this,  we solve this compound inequality using rules of inequality to get x all by itself.

 x > a   or  x < -a

\dpi{120} x \geq a  or  \dpi{120} x\leq -a

After breaking the inequalities like this, expression can be solved further to get x by itself.

 Since absolute value is always positive, it can’t be less than or equal to any negative number so this case has no solution. Since absolute value is always positive ,it would always be greater than any negative number so this case has ‘All real numbers ’ as solution.

 

Lets work on some examples representing all the 4 cases to understand them better.

 

Example1. Solve the following inequality

             \dpi{120} {\color{Red} \left | 2x+3 \right |\leq 15}

Solution: This is of the form   \mathbf{\left | x \right |}\leq \mathbf{a}  (case 1) , so rewriting the inequality we get,

\dpi{120} -15\leq 2x+3\leq 15

-3                  -3         -3            subtracting 3 from all sides, we get ,

\dpi{120} -18 \leq 2x\leq 12

\dpi{120} \frac{-18}{2}\leq \frac{2x}{2}\leq \frac{12}{2}

\dpi{120} -9\leq x\leq 6

In interval notation this solution can be written as   [-9,6]

 

Example2. Solve the following inequality and write the solution in interval notation.

           3|2x+1| > 9

Solution: First we isolate the absolute part. So divide both sides by 3 we get,          |2x+1| > 3

This is of the form   \mathbf{\left | x \right |}> \mathbf{a}  (case2) , so rewriting the inequality we get,

2x+1 > 3    or   2x+1 < -3

2x > 2                 2x  < -4

x > 1    or               x  < -2

Combining these two inequalities by replacing ‘or’ with U (union symbol) to  write final answer in interval notation we get,

\dpi{120} \left ( -\infty ,-2 \right )\cup \left ( 1,\infty \right )

 

Example3. Solve the following inequality and write the solution in interval notation.

       \dpi{120} 8+\left | 2x+5 \right |\leq 5{\color{Red} }

Solution: Subtract 8 from both sides to isolate absolute part  we get,

\dpi{120} \left | 2x+5 \right |\leq -3

This is of the form    \mathbf{\left | x \right |}\leq \mathbf{-a}  (case 3).

Since absolute value is always positive, so it can’t be less than or equal to any negative number. Therefore this inequality has  No Solution.

 

Example4. Solve the following inequality and write the solution in interval notation.

        \dpi{120} 8+\left | 2x+5 \right |>5

Solution: Subtract 8 from both sides to isolate absolute part  we get,

\dpi{120} \left | 2x+5 \right |> -3

This is of the form   \mathbf{\left | x \right |} > \mathbf{-a}  (case 4).

Since absolute value is always positive, so it is always greater than any negative number. Therefore this inequality has ‘All Real numbers ’ as solution which can be written in interval notation as, (-∞,∞)

 

Inequalities with absolute sign having variables on both sides:

Inequalities having more than one absolute sign  or having variable on both sides, are solved using different approach. We get the intervals on a number line according to expression inside absolute signs and then solve inequalities for each interval. Lets work on some examples to understand this process better.

 

Example5. Solve the given inequality and write the answer in interval notation.                            

             |x-1|+|x-2| ≥ 4

Solution:  First we set the expressions inside absolute signs =0 to get points on number line.

So            x-1=0   =>  x=1   and      x-2=0 => x=2

Using these two points, we get the intervals on number line as (-∞,1) ,(1,2) and (2,∞). Here we also make use of definition of absolute function.

\dpi{120} \left | x-1 \right |= \left\{\begin{matrix} -(x-1) & x < 1 \\ (x-1) &x\geq 1 \end{matrix}\right.                        \dpi{120} \left | x-2 \right |= \left\{\begin{matrix} -(x-2) & x < 2 \\ (x-2) &x\geq 2 \end{matrix}\right.

For interval (-∞,1) we have the inequality as,

-(x-1)-(x-2) ≥ 4

-x+1-x+2 ≥ 4

-2x +3 ≥ 4

-2x ≥ 1

x ≤ -½   —–(i)

For interval (1,2) we have the inequality as,

(x-1) –(x-2) ≥ 4

x-1-x+2 ≥ 4

1 ≥ 4  (which is absurd )

For interval (2,∞), we have the inequality as,

(x-1) +(x-2) ≥ 4

2x -3   ≥ 4

2x  ≥ 7

\dpi{120} x\geq \frac{7}{2}   ——–(ii)

Combining results (i) and (ii) we get the final solution in interval form as,           (-∞ , -½]  U [ 7/2 ,∞)

 

Example6. Solve the inequality  |2x-3| > x-1

Solution: Taking positive and negative cases separately,

-(2x-3) > x-1                or                           2x-3 > x-1

-2x+3 > x-1                                                2x-x > -1+3

-2x-x > -1-3                                                 x  > 2

-3x  > -4

x < 4/3

Combining these two inequalities, we get the final answer in interval form as,     (-∞, 4/3)  U (2,∞)

 

Practice problems:

Solve the following inequalities and write the solution in interval form.

1)   |3x-2| ≤ ½

2)   |4-x| +1 > 3

3)   |x-1|+|x-2|+|x-3|≥ 6

4)   4+|x-3| < 1

 

 

 

 

 

 

 

Answers:

1)   [1/2 , 5/6]

2)   (-∞,2) U(6,∞)

3)   (-∞,0] U [4,∞)

4)   No solution

Solving linear inequalities

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Solving linear inequalities is similar to solving linear equations. All the rules and facts are same as  for linear equation , only exception is that whenever we multiply or divide any inequality with a negative number, its sign get reversed. That means < sign get changed to >  and viceversa.

Here are few examples to understand the process.

Example1. Solve the following inequality.

2(x-3) > 10

Solution:           \dpi{120} \frac{2(x-3)}{2}> \frac{10}{2}

(x-3)  > 5

X  > 5+3

X > 8

Now  we work on same example but with a negative sign and see how it effect  the solution.

Example2. Solve the following inequality.

-2(x-3) > 10

Solution:  To  get rid of -2  , we divide both sides by -2 and therefore inequality sign get reversed.

\dpi{120} \frac{-2(x-3)}{-2}< \frac{10}{-2}

(x-3)  < -5

X  < -5+3   =>     x <  -2

 

Representing the solution of an inequality : Solution of an inequality can be represented in two ways. One  as an inequality  and other as interval notation. To represent the solution in interval notation we need to be familiar with some rules.

Rules to convert the inequality in interval notation:

1) In interval notation smaller number is always written towards left (first place) and larger number is written towards right(second place). This rule would be helpful in solving double inequalities.

2) Solution having only < or > sign is represented with open brackets (   ) which means end points are not included

3) The solution having  or  is represented with closed brackets [  ] which means end points are included.

4) We always use parenthesis (  ) with infinity.

 

Here is a beautiful tabular representation of all these rules.

Representing the solution of inequality on a number line :

Inequality Interval Notation Number line representation
a<x<b  (a,b)  
a≤x≤b [a,b)  
a<x≤ b (a,b]  
a≤x ≤b [a,b]  
X<b (-∞ ,b)  
x>a (a,\bg_white \large \mathbf{\infty} )  
x ≤ b (-\dpi{120} \infty, b]  
x ≥ a [a,\dpi{120} \infty )  

Example3. Solve the following inequality and represent the solution as interval notation.

          12-(2t+1) \dpi{100} \geq  2(t-5)

Solution:  First we remove parenthesis , for that we distribute – sign on left and 2 on right side of the inequality.

12 -2t-1  \dpi{120} \geq  2t-10

-2t -2t \dpi{120} \geq  -10-12+1          (combining and rearranging like terms)

-4t \dpi{120} \geq  -21

\dpi{120} \frac{-4t}{-4}\leq \frac{-21}{-4}                (inequality reversed when divided by negative number)

\dpi{120} t \leq \frac{-21}{4}

In interval notation answer can be written as,

\dpi{100} \left ( -\infty ,\frac{-21}{4} \right )

 

Example4. Solve the following inequality and write the answer in interval notation. Also represent it on number line.

\dpi{100} 7+\frac{1}{5}\left ( x-10 \right )\geq \left ( x+1 \right )

Solution:   Distributing 1/5 we get,

\dpi{100} 7+\frac{x}{5}-2\geq \ x+1

\dpi{100} \frac{x}{5}-x\geq 1-7+2

\dpi{100} \frac{-4x}{5}\geq -4

Multiply both sides with -5/4  to get x by itself,

\dpi{100} \frac{-5}{4}*\frac{-4x}{5}\leq -4*\frac{-5}{4}                  (inequality reversed when multiplied by negative number)

x  \dpi{120} \leq  5

Interval notation:   (- \dpi{120} \infty, 5]

How to solve Double inequality: Solving double inequality is similar to solving single inequality but here we don’t move the variable to one side and numbers to other side as we do in solving single inequality. Our target is to get x all by itself and to get that,we use same operation to both sides. That means if we add or subtract something to middle x term, we do the same operation to left and right side terms too.

 

Example5. Solve the following inequality and represent the solution  in interval notation.

7 < 5+2x < 1

Solution:    7 < 5+2x < 1

-5    -5        -5                (subtracting 5 from all sides)

2 <   2x  < -4

\dpi{120} \frac{2}{2}< \frac{2x}{2}< \frac{-4}{2}

1 < x < -2

As per rules of writing interval notations, smaller number should be written first, so this inequality can be rewritten as,

-2 < x < 1

Which can be further written in interval notation as (-2,1)

 

Example6. Solve the given compound inequality and represent the solution in interval notation. Also represent the solution on number line.                                -10 < -5(x+2) \dpi{120} \leq  6

Solution:             -10 < -5x-10 \dpi{120} \leq 6                    (distributed -5)\dpi{100} 7+\frac{1}{5}\left ( x-10 \right )\geq \left ( x+1 \right ){\color{Red} }

-10  <  -5x-10 \dpi{120} \leq   6

+10          +10     +10

0  <   -5x   \dpi{120} \leq     16

\dpi{120} \frac{0}{-5}> \frac{-5x}{-5}\geq \frac{16}{-5}

\dpi{100} 0> x\geq \frac{-16}{5}

Writing this solution in proper format , we get interval notation as

\dpi{100} [\frac{-16}{5},0)  => [-3.2 ,0)

 

 

Practice problems:

Solve the following inequalities and represent the solutions in interval notation.

  1. 3(2x-5) \dpi{120} \leq -15
  2. -2(x+1) > 8-(x+5)

Solve the following compound inequalities and represent the solution on number line.

  1.   -5 \dpi{120} \leq 2x-1 \dpi{120} \leq  3
  2.    9 \dpi{120} \geq 3(x-1) \dpi{120} \geq -12

 

 

 

Answers:

  • (-\dpi{120} \infty ,0]
  • (-\dpi{120} \infty , -5)
  • [-2,2]
  • [-3,4]