Question

How do you find the square root of 38.94?

Answer 1

Hint: Calculating the square roots perfect squares is very easy. The square root of other natural numbers can also be calculated easily by the process of grouping the factors. Here, we are given a decimal number to find the square root for. Thus, we will use the approximation method using differentials to calculate the square root of 38.94.

Complete step-by-step solution:
We have to calculate the value of $\sqrt{38.94}$. We will use the method of approximations using differentials. It is given as:
$f\left( x+\Delta x \right)=f\left( x \right)+{f}'\left( x \right).\Delta x$
Here, we have the function, $f\left( x \right)=\sqrt{x}$.
We shall differentiate this function with respect to x to find ${f}'\left( x \right)$. Thus differentiating $f\left( x \right)=\sqrt{x}$ with respect to x, we get
${f}'\left( x \right)=\dfrac{d}{dx}\sqrt{x}$
$\Rightarrow {f}'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
Also, here we must have $x+\Delta x=38.94$. Therefore, we take $x=36$ and $\Delta x=2.94$
Therefore, on substituting the values, the equation modifies to
$f\left( 36+2.94 \right)=f\left( 36 \right)+{f}'\left( 36 \right).\left( 2.94 \right)$
$\Rightarrow f\left( 38.94 \right)=f\left( 36 \right)+{f}'\left( 36 \right).\left( 2.94 \right)$
Further putting the values of the function $f\left( x \right)$, we get
$\Rightarrow \sqrt{38.94}=\sqrt{36}+\dfrac{1}{2\sqrt{36}}.\left( 2.94 \right)$
We know that, $6\times 6=36$. This implies that $\sqrt{36}=6$
$\begin{align}
  & \Rightarrow \sqrt{38.94}=6+\dfrac{1}{2\left( 6 \right)}.\left( 2.94 \right) \\
 & \Rightarrow \sqrt{38.94}=6+\dfrac{1}{6}.\left( 1.47 \right) \\
\end{align}$
 Taking LCM equals 6 on the right-hand side of equation, we get
$\begin{align}
  & \Rightarrow \sqrt{38.94}=\dfrac{36+1.47}{6} \\
 & \Rightarrow \sqrt{38.94}=\dfrac{37.47}{6} \\
\end{align}$
$\Rightarrow \sqrt{38.94}=6.245$
Therefore, the square root of 38.94 is 6.245.

Note: While solving problems using the approximations method, the given number is divided into two parts, $x$ and $\Delta x$. To calculate the square root, we must keep in mind that x should be chosen such that it is the closest perfect square to the given number. It can be bigger or smaller than the number given but it should be the closest perfect square to it on the number line. This makes $\Delta x$ very small and gives us the most accurate value.