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How do you evaluate $2\sqrt{44{{x}^{3}}}-\sqrt{7}-\sqrt{99{{x}^{3}}}+\sqrt{63}$?
Answer
434.1k+ views
Hint: To evaluate the given expression $2\sqrt{44{{x}^{3}}}-\sqrt{7}-\sqrt{99{{x}^{3}}}+\sqrt{63}$, we are going to take some common terms from $2\sqrt{44{{x}^{3}}}-\sqrt{99{{x}^{3}}}$ and also some common terms from $-\sqrt{7}+\sqrt{63}$ and then we will simplify further to get the final answer.
Complete step by step solution:
The expression given in the above problem is as follows:
$2\sqrt{44{{x}^{3}}}-\sqrt{7}-\sqrt{99{{x}^{3}}}+\sqrt{63}$
Now, writing $2\sqrt{44{{x}^{3}}}$ and $-\sqrt{99{{x}^{3}}}$ together along with that we are also going to write $-\sqrt{7}$ and $\sqrt{63}$ together in the above expression and we get,
$\Rightarrow 2\sqrt{44{{x}^{3}}}-\sqrt{99{{x}^{3}}}-\sqrt{7}+\sqrt{63}$
In the above expression, we are going to write 44 as $11\times 4$ and 99 as $11\times 9$ and 63 as $7\times 9$ and we get,
$\Rightarrow 2\sqrt{11\times 4{{x}^{3}}}-\sqrt{11\times 9{{x}^{3}}}-\sqrt{7}+\sqrt{7\times 9}$
Taking $\sqrt{11{{x}^{3}}}$ as common from the first two terms and $\sqrt{7}$ from the last two terms we get,
$\Rightarrow \sqrt{11{{x}^{3}}}\left( 2\sqrt{4}-\sqrt{9} \right)-\sqrt{7}\left( 1-\sqrt{9} \right)$
In the above expression, we can write $\sqrt{4}$ as 2 and $\sqrt{9}$ as 3 in the above expression and we get,
$\begin{align}
& \Rightarrow \sqrt{11{{x}^{3}}}\left( 2\left( 2 \right)-3 \right)-\sqrt{7}\left( 1-3 \right) \\
& \Rightarrow \sqrt{11{{x}^{3}}}\left( 4-3 \right)-\sqrt{7}\left( -2 \right) \\
\end{align}$
We know that multiplying two negatives will become positive so using this property in the above and we get,
$\Rightarrow \sqrt{11{{x}^{3}}}+2\sqrt{7}$
Hence, we have simplified the above expression to $\sqrt{11{{x}^{3}}}+2\sqrt{7}$.
Note: You might think how we have taken only those terms as common. The reason is if you look carefully the given expression you will find that:
$2\sqrt{44{{x}^{3}}}-\sqrt{7}-\sqrt{99{{x}^{3}}}+\sqrt{63}$
If you see the first and third term then you will see $\sqrt{11{{x}^{3}}}$ is a common term and in the second and last term, we can take $\sqrt{7}$ as common. Now, this observation will come when you have a good command on the multiples of 1, 2, 3, 4……. so on. This observation will get better when you practice a lot of questions then by just looking at the problem you will know which numbers to take as common.
Complete step by step solution:
The expression given in the above problem is as follows:
$2\sqrt{44{{x}^{3}}}-\sqrt{7}-\sqrt{99{{x}^{3}}}+\sqrt{63}$
Now, writing $2\sqrt{44{{x}^{3}}}$ and $-\sqrt{99{{x}^{3}}}$ together along with that we are also going to write $-\sqrt{7}$ and $\sqrt{63}$ together in the above expression and we get,
$\Rightarrow 2\sqrt{44{{x}^{3}}}-\sqrt{99{{x}^{3}}}-\sqrt{7}+\sqrt{63}$
In the above expression, we are going to write 44 as $11\times 4$ and 99 as $11\times 9$ and 63 as $7\times 9$ and we get,
$\Rightarrow 2\sqrt{11\times 4{{x}^{3}}}-\sqrt{11\times 9{{x}^{3}}}-\sqrt{7}+\sqrt{7\times 9}$
Taking $\sqrt{11{{x}^{3}}}$ as common from the first two terms and $\sqrt{7}$ from the last two terms we get,
$\Rightarrow \sqrt{11{{x}^{3}}}\left( 2\sqrt{4}-\sqrt{9} \right)-\sqrt{7}\left( 1-\sqrt{9} \right)$
In the above expression, we can write $\sqrt{4}$ as 2 and $\sqrt{9}$ as 3 in the above expression and we get,
$\begin{align}
& \Rightarrow \sqrt{11{{x}^{3}}}\left( 2\left( 2 \right)-3 \right)-\sqrt{7}\left( 1-3 \right) \\
& \Rightarrow \sqrt{11{{x}^{3}}}\left( 4-3 \right)-\sqrt{7}\left( -2 \right) \\
\end{align}$
We know that multiplying two negatives will become positive so using this property in the above and we get,
$\Rightarrow \sqrt{11{{x}^{3}}}+2\sqrt{7}$
Hence, we have simplified the above expression to $\sqrt{11{{x}^{3}}}+2\sqrt{7}$.
Note: You might think how we have taken only those terms as common. The reason is if you look carefully the given expression you will find that:
$2\sqrt{44{{x}^{3}}}-\sqrt{7}-\sqrt{99{{x}^{3}}}+\sqrt{63}$
If you see the first and third term then you will see $\sqrt{11{{x}^{3}}}$ is a common term and in the second and last term, we can take $\sqrt{7}$ as common. Now, this observation will come when you have a good command on the multiples of 1, 2, 3, 4……. so on. This observation will get better when you practice a lot of questions then by just looking at the problem you will know which numbers to take as common.
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