Answer
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Hint: We first try to form the imaginary numbers for $-59$ using the identities ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. Then we get the root value in the form of both positive and negative values and imaginary numbers. We also find the decimal value for $\sqrt{59}$.
Complete step-by-step answer:
The square root value of the negative value gives imaginary values.
Therefore, we use the know identity values and relations for imaginary $i=\sqrt{-1}$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
Therefore, $-59=\left( -1 \right)\times 59=59{{i}^{2}}$.
Now taking the root value we get $\sqrt{-59}=\sqrt{59{{i}^{2}}}=\pm i\sqrt{59}$.
We can also find the root value in decimal for 59 as 59 is a prime number.
We take 2 digits as a set from the right end and complete the division. For decimal form we take the set from the right side of the decimal.
\[\begin{align}
& 7 \\
& 7\left| \!{\overline {\,
\begin{align}
& \overline{59}.\overline{00}\overline{00} \\
& \underline{49} \\
& 10.00 \\
\end{align} \,}} \right. \\
\end{align}\]
Now we have to enter the decimal part. We keep doing the breaking in the set form till 3-digit place after decimal.
\[\begin{align}
& 7.67 \\
& 146\left| \!{\overline {\,
\begin{align}
& 1000\overline{00} \\
& \underline{876} \\
& 12400 \\
\end{align} \,}} \right. \\
& 1527\left| \!{\overline {\,
\begin{align}
& 12400 \\
& \underline{10689} \\
& 1711 \\
\end{align} \,}} \right. \\
\end{align}\]
So, $\sqrt{-59}=\pm i\sqrt{59}=\pm 7.67i$
Therefore, the square root of $-59$ is $\pm i\sqrt{59}=\pm 7.67i$.
So, the correct answer is “$\pm 7.67i$”.
Note: The long-division method and arranging the set of 2 digits is different for integer and decimal. But taking double for the next division and putting a particular number is the same process for both of them. Since 3 is a non-perfect square number, we will find the value of root 3 using the long division method as shown above.
Complete step-by-step answer:
The square root value of the negative value gives imaginary values.
Therefore, we use the know identity values and relations for imaginary $i=\sqrt{-1}$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
Therefore, $-59=\left( -1 \right)\times 59=59{{i}^{2}}$.
Now taking the root value we get $\sqrt{-59}=\sqrt{59{{i}^{2}}}=\pm i\sqrt{59}$.
We can also find the root value in decimal for 59 as 59 is a prime number.
We take 2 digits as a set from the right end and complete the division. For decimal form we take the set from the right side of the decimal.
\[\begin{align}
& 7 \\
& 7\left| \!{\overline {\,
\begin{align}
& \overline{59}.\overline{00}\overline{00} \\
& \underline{49} \\
& 10.00 \\
\end{align} \,}} \right. \\
\end{align}\]
Now we have to enter the decimal part. We keep doing the breaking in the set form till 3-digit place after decimal.
\[\begin{align}
& 7.67 \\
& 146\left| \!{\overline {\,
\begin{align}
& 1000\overline{00} \\
& \underline{876} \\
& 12400 \\
\end{align} \,}} \right. \\
& 1527\left| \!{\overline {\,
\begin{align}
& 12400 \\
& \underline{10689} \\
& 1711 \\
\end{align} \,}} \right. \\
\end{align}\]
So, $\sqrt{-59}=\pm i\sqrt{59}=\pm 7.67i$
Therefore, the square root of $-59$ is $\pm i\sqrt{59}=\pm 7.67i$.
So, the correct answer is “$\pm 7.67i$”.
Note: The long-division method and arranging the set of 2 digits is different for integer and decimal. But taking double for the next division and putting a particular number is the same process for both of them. Since 3 is a non-perfect square number, we will find the value of root 3 using the long division method as shown above.
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