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# How do you find the square root of $- 59$?

Last updated date: 22nd Jun 2024
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Hint: Here we are given to find the square root of the negative number which means that we need to find that number which on multiplied with itself will give us the result as $- 59$.
Now there cannot be any real number that will give us the resultant as $- 59$ and therefore we can say that this will be an imaginary number. So we can write $- 59$ as $\left( {59} \right)\left( {{i^2}} \right)$ and here $i$ represents iota and its value is $\sqrt { - 1}$.

Complete step by step solution:
Here we are given to find the square root of $- 59$ and we know that all real numbers have the square as the positive numbers but here we are given the negative number as the square of the number. Hence we come to know that this cannot be a real number but it will be an imaginary number. In the imaginary number we must know that $i$ represents iota and its value is $\sqrt { - 1}$.
So we can say that $i = \sqrt { - 1}$
Hence we can square both the sides and get:
${i^2} = - 1$
Hence we can write $- 59$ as $\left( {59} \right)\left( {{i^2}} \right)$
Now writing this in the square root form, we will get:
$\sqrt {\left( {59} \right)\left( {{i^2}} \right)}$
Now we know that when we find the square root of any number we need to write the number in the form of factors and then write the number out of the root whose pair we have in the root.
Hence we can say:
$\sqrt {\left( {59} \right)\left( {{i^2}} \right)} = \sqrt {\left( {59} \right)\left( i \right)\left( i \right)}$
As we can see that iota is in pairs so we can take it outside.
$\sqrt {\left( {59} \right)\left( {{i^2}} \right)} = \sqrt {\left( {59} \right)\left( i \right)\left( i \right)} = i\sqrt {\left( {59} \right)}$

Hence we get the result as $i\sqrt {\left( {59} \right)}$

Note:
Here the student must know that whenever we are given to find the square root of any negative number, then we will always get the imaginary number as our answer because no real number can have the square as negative.