Question

How do you simplify $\sqrt{-1256}$?

Answer 1

Hint: In the above question, we have been given a square root expression to be simplified. To simplify it, we need to use the law of radicals which is given by $\sqrt{ab}=\sqrt{a}\sqrt{b}$ so that the given expression reduces to $\sqrt{-1}\sqrt{1256}$. Then, since $i=\sqrt{-1}$, the given expression will become $\sqrt{1256}i$. Then we need to simplify the expression $\sqrt{1256}$ by writing the prime factorisation of the number $1256$. From the prime factorisation, we can take out the highest perfect square from the square root to get the final simplified expression.

Complete step-by-step answer:
The expression given in the above question is
$\Rightarrow \sqrt{-1256}$
Now, using the law of radicals given by $\sqrt{ab}=\sqrt{a}\sqrt{b}$, we can write the above expression as
$\Rightarrow \sqrt{-1}\sqrt{1256}$
Now, we know that $\sqrt{-1}$ is a complex number which is equal to $i$ so that we can put $\sqrt{-1}=1$ in the above expression to get
$\Rightarrow \sqrt{1256}i.......\left( i \right)$
Now, in order to further simplify the above 0065pression, we have to write the prime factorization for the number $1256$, as shown below.
\[\begin{align}
  & 2\left| \!{\underline {\,
  1256 \,}} \right. \\
 & 2\left| \!{\underline {\,
  628 \,}} \right. \\
 & 2\left| \!{\underline {\,
  314 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  157 \,}} \right. \\
\end{align}\]
From the above, we got the prime factorization of the number $1256$ as
\[\begin{align}
  & \Rightarrow 1256=2\times 2\times 2\times 157 \\
 & \Rightarrow 1256=4\times 314 \\
\end{align}\]
Substituting this in (i) we get
$\Rightarrow \sqrt{4\times 314}i$
Again using the law of radicals given by $\sqrt{ab}=\sqrt{a}\sqrt{b}$, we can write the above expression as
\[\Rightarrow \sqrt{4}\times \sqrt{314}i\]
Now, we know that \[\sqrt{4}=2\]. Putting this above we get
$\Rightarrow 2\sqrt{314}i$
Hence, the given expression has been finally simplified to $2\sqrt{314}i$.

Note: We might argue as to why we haven’t substituted $\sqrt{4}=\pm 2$ to get two simplified expressions, one positive and the other negative. The answer to this question is that the square root of a number is always positive. If we have the equation ${{x}^{2}}=4$, then the solution to this equation will be both $2$ and $-2$. But the value of $\sqrt{4}$ is $2$ only.