Question

If A and G be A.M And G.M, respectively between two positive numbers, prove that the numbers are $A\pm \sqrt{\left( A+G \right)\left( A-G \right)}$.

Answer 1

Hint: We know that A.M is Arithmetic Mean and G.M is Geometric Mean and the A.M and G.M for two positive 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 -
We have been given A and G as A.M and G.M between two positive numbers and we have to prove that the numbers are $A\pm \sqrt{\left( A+G \right)\left( A-G \right)}$. To solve this question, let us start by assuming the two positive numbers as a and b. We know that A.M between two positive numbers is the average of the two numbers. So, it is given as,
$A=\dfrac{a+b}{2}\ldots \ldots \ldots \left( 1 \right)$
We also know that the G.M between any two numbers is the square root of the product of the numbers. So, it is given as,
$G=\sqrt{ab}$
On squaring both sides of the above equality, we will get,
$\begin{align}
  & {{G}^{2}}=ab \\
 & \Rightarrow \dfrac{{{G}^{2}}}{a}=b\ldots \ldots \ldots \left( 2 \right) \\
\end{align}$
We will now substitute the value of b in equation (1). So, we get,
$\begin{align}
  & A=\dfrac{a+\dfrac{{{G}^{2}}}{a}}{2} \\
 & \Rightarrow 2A=\dfrac{{{a}^{2}}+{{G}^{2}}}{a} \\
 & \Rightarrow 2Aa={{a}^{2}}+{{G}^{2}} \\
\end{align}$
By rearranging the terms, we get,
${{a}^{2}}-2Aa+{{G}^{2}}=0$
We know that the roots of a quadratic equation $a{{x}^{2}}+bx+c$ is given by, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, we will apply this to the above equation, where we have $x=a,a=1,b=-2A,c={{G}^{2}}$. So, we get,
$\begin{align}
  & a=\dfrac{-\left( -2A \right)\pm \sqrt{{{\left( -2A \right)}^{2}}-4{{G}^{2}}}}{2} \\
 & \Rightarrow a=\dfrac{2A\pm \sqrt{4{{A}^{2}}-4{{G}^{2}}}}{2} \\
 & \Rightarrow a=\dfrac{2A\pm 2\sqrt{{{A}^{2}}-{{G}^{2}}}}{2} \\
 & \Rightarrow a=2\left( \dfrac{A\pm \sqrt{{{A}^{2}}-{{G}^{2}}}}{2} \right) \\
 & \Rightarrow a=A\pm \sqrt{{{A}^{2}}-{{G}^{2}}} \\
\end{align}$
Now we will substitute $a=A+\sqrt{{{A}^{2}}-{{G}^{2}}}$ in equation (1). So, we get,
$\begin{align}
  & A=\dfrac{A+\sqrt{{{A}^{2}}-{{G}^{2}}}+b}{2} \\
 & \Rightarrow 2A=A+\sqrt{{{A}^{2}}-{{G}^{2}}}+b \\
 & \Rightarrow 2A-A-\sqrt{{{A}^{2}}-{{G}^{2}}}=b \\
 & \Rightarrow b=A-\sqrt{{{A}^{2}}-{{G}^{2}}} \\
\end{align}$
We will now use the identity, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above equality. So, we get,
$b=A-\sqrt{\left( A-G \right)\left( A+G \right)}$
Now, we will substitute $a=A-\sqrt{{{A}^{2}}-{{G}^{2}}}$ in equation (1). So, we get,
$\begin{align}
  & A=\dfrac{A-\sqrt{{{A}^{2}}-{{G}^{2}}}+b}{2} \\
 & \Rightarrow 2A=A-\sqrt{{{A}^{2}}-{{G}^{2}}}+b \\
 & \Rightarrow 2A-A+\sqrt{{{A}^{2}}-{{G}^{2}}}=b \\
 & \Rightarrow b=A+\sqrt{{{A}^{2}}-{{G}^{2}}} \\
\end{align}$
We will again use the identity, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. So, we get,
$b=A+\sqrt{\left( A-G \right)\left( A+G \right)}$
Hence, we get the two numbers as $A+\sqrt{\left( A-G \right)\left( A+G \right)}$ and $A-\sqrt{\left( A-G \right)\left( A+G \right)}$. Therefore, it is proved.

Note: We have to be careful while solving quadratic equation questions as there are chances of mistakes with the signs while finding the roots. Sometimes the students try to use factorisation 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. You can also remember the statement to be proved, given in the question as a property for two positive numbers.