Question

If a > b > 0 are two real numbers, then the value of $\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+.....}}}}$ is?
(a) Independent of b
(b) Independent of a
(c) Independent of both a and b
(d) Dependent on both a and b

Answer 1

Hint: Assume the given expression as x. Now, write the expression of the sum of infinite terms as$x=\sqrt{ab+\left( a-b \right)x}$. Square both the sides and take all the terms to the L.H.S to form a quadratic expression in x. Find the roots of the equation by using the discriminant formula of a quadratic equation given as $x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$ where A is the coefficient of ${{x}^{2}}$, B is the coefficient of x and C is the constant term. Reject the negative value of x to get the correct option.

Complete step by step answer:
Here we have been provided with the expression $\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+.....}}}}$ with the condition a > b > 0 and we are asked to find its value and choose the correct option. Let us assume the value of this expression as x, so we have,
$\Rightarrow x=\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+.....}}}}$
As we can see that there are infinite terms in the given expression so we cannot add them directly. Now, we can see that at a certain step the expression is repeating itself so we can write it as: -
$\Rightarrow x=\sqrt{ab+\left( a-b \right)x}$
On squaring both the sides we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}=ab+\left( a-b \right)x \\
 & \Rightarrow {{x}^{2}}-\left( a-b \right)x-ab=0 \\
\end{align}$
The above equation is a quadratic equation so using the discriminant formula given as $x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$ where A is the coefficient of ${{x}^{2}}$, B is the coefficient of x and C is the constant term we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left[ -\left( a-b \right) \right]\pm \sqrt{{{\left( a-b \right)}^{2}}-4\left( 1 \right)\left( -ab \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{\left( a-b \right)\pm \sqrt{{{\left( a-b \right)}^{2}}+4ab}}{2} \\
\end{align}\]
Using the algebraic identity ${{\left( a-b \right)}^{2}}+4ab={{\left( a+b \right)}^{2}}$ we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( a-b \right)\pm \sqrt{{{\left( a+b \right)}^{2}}}}{2} \\
 & \Rightarrow x=\dfrac{\left( a-b \right)\pm \left( a+b \right)}{2} \\
\end{align}\]
(1) Considering the positive sign we have,
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( a-b \right)+\left( a+b \right)}{2} \\
 & \Rightarrow x=\dfrac{2a}{2} \\
 & \Rightarrow x=a \\
\end{align}\]
(2) Considering the negative sign we have,
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( a-b \right)-\left( a+b \right)}{2} \\
 & \Rightarrow x=\dfrac{-2b}{2} \\
 & \Rightarrow x=-b \\
\end{align}\]
Now, the value of the expression $\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+\left( a-b \right)\sqrt{ab+.....}}}}$ will always be positive because the term $ab>0$ and $\left( a-b \right)>0$ due of the condition a > b > 0 given in the question. Also the square root is a positive square root. Therefore the value x = –b in invalid because it is less than 0, so the valid value of the expression is x = a, which is independent of b but depends on a.

So, the correct answer is “Option a”.

Note: You must be careful about the condition given in the question which is a > b > 0 because if you neglect it then you will get two values of x and will get confused in the four options. Note that in case we have to consider the negative square roots also then the minus sign is provided in the question and we don’t have to consider it ourselves.