Question

How do you simplify $4\sqrt{7}+8\sqrt{63}$

Answer 1

Hint: Now to simplify the given expression we will first write 63 as a product of 9 and 3.Now we know that $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ using this we will simplify the expression and write the square root of 9. Now using distributive property we know that $\left( a+b \right)c=ac+bc$ hence we will simplify the given expression.

Complete step by step solution:
Now let us understand the concept of square roots and squares.
Square of a number means the number with power 2.
Hence ${{a}^{2}}$ is known as square of a.
Now square root can be explained as the inverse function of square. Square root of a is denoted by $\sqrt{a}$
Hence we have $\sqrt{{{a}^{2}}}=a$
Now let us understand this with an example.
We know that ${{3}^{2}}=3\times 3=9$ hence square of 3 is 9.
Similarly we have $\sqrt{9}=\sqrt{{{3}^{2}}}=3$ hence the square root of 9 is 3.
Now we cannot add or subtract the terms in square root directly.
Hence $\sqrt{a}+\sqrt{b}\ne \sqrt{a+b}$ . Though we can take the square root of a number common through distributive property.
To multiply the terms in square root we have $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$
Now consider the given expression $4\sqrt{7}+8\sqrt{63}$ .
Now we know that 63 can be written as $9\times 7$ .
Hence we have,
$\begin{align}
  & \Rightarrow 4\sqrt{7}+8\sqrt{9\times 7} \\
 & \Rightarrow 4\sqrt{7}+8\left( \sqrt{9}\times \sqrt{7} \right) \\
\end{align}$
Now we know that square root of 9 is 3 hence we get,
$\begin{align}
  & \Rightarrow 4\sqrt{7}+8\left( 3\sqrt{7} \right) \\
 & \Rightarrow 4\sqrt{7}+24\sqrt{7} \\
\end{align}$
Now we know according to distributive property $\left( a+b \right)c=ac+bc$. Hence using this we get,
$\begin{align}
  & \Rightarrow \left( 4+24 \right)\sqrt{7} \\
 & \Rightarrow 28\sqrt{7} \\
\end{align}$
Hence the value of the given expression is $28\sqrt{7}$ .

Note:
Now we know if we have 2 in the power of a term then it is called square. Similarly if 3 is in power we call it cube. Now for cubes the inverse function is called cube root and is denoted by $\sqrt[3]{\left( x \right)}$ . Now in general the ${{n}^{th}}$ root of a number is denoted by $\sqrt[n]{x}$ and we have $\sqrt[n]{{{x}^{n}}}=x$