Question

Simplify $3\sqrt 3 + 2\sqrt {27} + \dfrac{7}{{\sqrt 3 }}$.

Answer 1

Hint: In this question we have to simplify the given expression. We can see that we have terms under root, so we will first try to break down the big numbers and try to put it in simpler form. After that we will try to remove the under root from the denominator by rationalising the given value. After this we will add the values.

Complete step by step solution:
Here we have,
$3\sqrt 3 + 2\sqrt {27} + \dfrac{7}{{\sqrt 3 }}$ .
We can break down the factors of $\sqrt {27} $ and write it in simpler value i.e.
$\sqrt {3 \times 3 \times 3} $
So it gives us value
 $\sqrt {{3^2} \times 3} = 3\sqrt 3 $
We will now rationalise the third part of the expression i.e.
$\dfrac{7}{{\sqrt 3 }}$
We will multiply the numerator and denominator with $\sqrt 3 $ , to get the value:
$\dfrac{7}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$
It gives the value
$\dfrac{{7\sqrt 3 }}{3}$ .
Now we put all the values back in the expression, and we have
 $3\sqrt 3 + 2\left( {3\sqrt 3 } \right) + \dfrac{{7\sqrt 3 }}{3}$
It can be written as
$3\sqrt 3 + 6\sqrt 3 + \dfrac{{7\sqrt 3 }}{3}$
We will take the common factor out and put the terms together i.e.
$\left( {3 + 6 + \dfrac{7}{3}} \right)\sqrt 3 $
We will now simplify this:
$\left( {\dfrac{{9 + 18 + 7}}{3}} \right)\sqrt 3 = \dfrac{{34}}{3}\sqrt 3 $
Hence the required answer is $\dfrac{{34\sqrt 3 }}{3}$.

Note:
We should note that we can also put the value of $\sqrt 3 = 1.732$ in the solution.
 It will give us
$\dfrac{{34 \times 1.732}}{3}$ .
On simplify we have
$\dfrac{{58.888}}{3} = 19.63(approx)$ .
We should know that rationalisation is done to make the given term simplified. It is done by multiplying and dividing the given term obtained on reversing the sign of the coefficient of under root of the denominator. Or we can say that root rationalisation is a process by which radicals in the denominator of an algebraic function are eliminated. The symbol of radical is $\left( {\sqrt {} } \right)$ .