
If A be the A.M. and G be the G.M. between two numbers. Show that the numbers are given by \[A \pm \sqrt {{A^2} - {G^2}} \]
Answer
518.7k+ views
Hint: We know that A.M. is arithmetic mean, and G.M. is geometric mean, and A.M and G.M. for two numbers say ‘a’ and ‘b’ will be \[\dfrac{{a + b}}{2}\] and \[\sqrt[{}]{{ab}}\] respectively. We will make a quadratic equation using it, and the roots of the equation gives the value of the numbers which we have to prove.
Complete step-by-step solution:
Given, A and G are A.M and G.M between two numbers. Let the two numbers be ‘a’ and ‘b’, we know that A.M between two numbers is the average of two numbers and G.M between two numbers is the square root of the product of the numbers.
Then, \[A = \dfrac{{a + b}}{2}\] and \[G = \sqrt {ab} \]
Simplifying them, we get:
\[ \Rightarrow a + b = 2A\,\,\,\,\,---------- equation\,1\]
and,
\[ \Rightarrow ab = {G^2}\,\,\,\,\, -------- equation\,2\]
\[\Rightarrow {\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab \\
\Rightarrow {\left( {a - b} \right)^2} = {\left( {2A} \right)^2} - 4{G^2} \\
\Rightarrow {\left( {a - b} \right)^2} = 4\left( {{A^2} - {G^2}} \right) \\
\Rightarrow \left( {a - b} \right) = \pm 2\sqrt {{A^2} - {G^2}} \,\,--- equation\,3\]
Taking equation 1 and equation 2 we get,
\[
\Rightarrow a - b = 2\sqrt {{A^2} - {G^2}} \\
\Rightarrow a + b = 2A \]
Now adding both we get:
\[ \Rightarrow a = A + \sqrt {{A^2} - {G^2}} \]
Putting this value of ‘a’ in equation 1, we get:
\[ \Rightarrow a + b = 2A \\
\Rightarrow A + \sqrt {{A^2} - {G^2}} + b = 2A \]
Calculating the value of ‘b’, we get:
\[ \Rightarrow b = A - \sqrt {{A^2} - {G^2}} \]
This equation can also we written as:
\[ = A - \sqrt {\left( {A + G} \right)\left( {A - G} \right)} \]
Hence, the numbers are \[A \pm \sqrt {{A^2} - {G^2}} \]
Note: We have to be careful while solving quadratic equation questions as there are chances of mistakes with signs while finding the roots. Sometimes the students try to use factorization methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving.
Complete step-by-step solution:
Given, A and G are A.M and G.M between two numbers. Let the two numbers be ‘a’ and ‘b’, we know that A.M between two numbers is the average of two numbers and G.M between two numbers is the square root of the product of the numbers.
Then, \[A = \dfrac{{a + b}}{2}\] and \[G = \sqrt {ab} \]
Simplifying them, we get:
\[ \Rightarrow a + b = 2A\,\,\,\,\,---------- equation\,1\]
and,
\[ \Rightarrow ab = {G^2}\,\,\,\,\, -------- equation\,2\]
\[\Rightarrow {\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab \\
\Rightarrow {\left( {a - b} \right)^2} = {\left( {2A} \right)^2} - 4{G^2} \\
\Rightarrow {\left( {a - b} \right)^2} = 4\left( {{A^2} - {G^2}} \right) \\
\Rightarrow \left( {a - b} \right) = \pm 2\sqrt {{A^2} - {G^2}} \,\,--- equation\,3\]
Taking equation 1 and equation 2 we get,
\[
\Rightarrow a - b = 2\sqrt {{A^2} - {G^2}} \\
\Rightarrow a + b = 2A \]
Now adding both we get:
\[ \Rightarrow a = A + \sqrt {{A^2} - {G^2}} \]
Putting this value of ‘a’ in equation 1, we get:
\[ \Rightarrow a + b = 2A \\
\Rightarrow A + \sqrt {{A^2} - {G^2}} + b = 2A \]
Calculating the value of ‘b’, we get:
\[ \Rightarrow b = A - \sqrt {{A^2} - {G^2}} \]
This equation can also we written as:
\[ = A - \sqrt {\left( {A + G} \right)\left( {A - G} \right)} \]
Hence, the numbers are \[A \pm \sqrt {{A^2} - {G^2}} \]
Note: We have to be careful while solving quadratic equation questions as there are chances of mistakes with signs while finding the roots. Sometimes the students try to use factorization methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

