Question

What is \[\sqrt {24} - \sqrt {54} + \sqrt {96} \] ?

Answer 1

Hint: Here we can simplify this numerical root by finding the perfect squares of each term and then eliminating the square root. And then, we need to simplify the term.

Complete step by step solution:
When we are need to simplify numerical root, we should look for these two key facts:
$\sqrt {{a^2}} = a$
$\sqrt {ab} = \sqrt a \sqrt b $
So, anytime we have a number inside a root , we should try to write it as a product of other numbers, of which at least one is a perfect square. Let’s analyze our case.
We can start the expression looks like this \[\sqrt {24} - \sqrt {54} + \sqrt {96} \]------ (1)
To try and simplify this expression, write out each value you have under a square root as a product of its prime factors, this will get,
From the first term of our expression
$\sqrt {24} = {2^3} \times 3 = {2^2} \times 2 \times 3$ and next,
$\sqrt {54} = 2 \times {3^3} = 2 \times 3 \times {3^2}$
$\sqrt {96} = {2^5} \times 3 = {2^4} \times 2 \times 3$
Notice that each number can be written as the product between a perfect square and $6$. This means that we can write,
$\sqrt {24} = \sqrt {{2^2} \times 6} $
$\sqrt {54} = \sqrt {{3^2} \times 6} $
$\sqrt {96} = \sqrt {{2^4} \times 6} $
Using the above mentioned properties, we get,
$\sqrt {24} = 2\sqrt 6 $
$\sqrt {54} = 3\sqrt 6 $
$\sqrt {96} = 4\sqrt 6 $
On substituting these values in equation (1) we get,
\[\sqrt {24} - \sqrt {54} + \sqrt {96} = 2\sqrt 6 - 3\sqrt 6 + 4\sqrt 6 \]
\[\sqrt {24} - \sqrt {54} + \sqrt {96} = \left( {2 - 3 + 4} \right)\sqrt 6 \]
\[\sqrt {24} - \sqrt {54} + \sqrt {96} = 3\sqrt 6 \]

The final answer is $3\sqrt 6 $.

Note: To simplify the terms inside of the radicals, try to factor them to find at least one term that is a perfect square. Once you do that, then you can take the square root of the perfect square and write it outside the radical, leaving the remaining factor inside the radical. The numbers outside the radical sign are the coefficients and the numbers inside it are the radicands.