# How would you simplify sqrt48 + sqrt3?

Feb 13, 2016

$5 \sqrt{3}$

#### Explanation:

We can split up $\sqrt{48}$ using the rule that $\sqrt{a b} = \sqrt{a} \sqrt{b}$.

Thus, $\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \sqrt{3} = 4 \sqrt{3}$.

The original expression can be rewritten as

$4 \sqrt{3} + \sqrt{3} = \sqrt{3} \left(4 + 1\right) = 5 \sqrt{3}$

Feb 13, 2016

Short story, try watching the result of division of the two numbers and substitute it in the bigger one.

$\sqrt{48} + \sqrt{3} = 5 \sqrt{3}$

#### Explanation:

Always try reducing the high number to something of which you know the root. Here someone should notice that:

$\frac{48}{3} = 16 \iff 48 = 3 \cdot 16$

The root of 16 is known, while the root of 3 can be factored. Therefore:

$\sqrt{48} + \sqrt{3}$

$\sqrt{16 \cdot 3} + \sqrt{3}$

$\sqrt{16} \cdot \sqrt{3} + \sqrt{3}$

$4 \cdot \sqrt{3} + \sqrt{3}$

$5 \sqrt{3}$

Feb 13, 2016

$5 \sqrt{3}$

#### Explanation:

radicals in simplified form are $a \sqrt{b}$
where a is a rational number.

to begin with , simplify $\sqrt{48}$

by considering the factors of 48 , particularly 'squares'

the factors required here are 16 (square) and 3.

using the following : $\sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a} b$

$\sqrt{48} = \sqrt{16} \times \sqrt{3} = 4 \sqrt{3}$

hence $\sqrt{48} + \sqrt{3} = 4 \sqrt{3} + \sqrt{3} = 5 \sqrt{3}$

Feb 13, 2016

$\textcolor{b l u e}{5 \sqrt{3}}$

:)

#### Explanation:

To simplify radicals, we must find first its largest "perfect squares" that evenly divides to simplify.

Since, $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$ (Theorem)

$\sqrt{48} + \sqrt{3}$

PerfectSquares:

$\textcolor{b l u e}{4}$ $= 2 \cdot 2$
$\textcolor{b l u e}{9}$ $= 3 \cdot 3$
$\textcolor{b l u e}{16}$ $= 4 \cdot 4$
$\textcolor{b l u e}{25}$ $= 5 \cdot 5$
$\textcolor{b l u e}{36}$ $= 6 \cdot 6$
$\textcolor{b l u e}{49}$ $= 7 \cdot 7$
$\textcolor{b l u e}{64}$ $= 8 \cdot 8$
$\textcolor{b l u e}{81}$ $= 9 \cdot 9$
$\textcolor{b l u e}{100}$ $= 10 \cdot 10$

we can simplify $\sqrt{48}$, as 48 evenly divides with $16$,

we get,

$\sqrt{48} = \sqrt{16 \cdot 3}$

using theorem from above,

$\sqrt{48} = \sqrt{16} \cdot \sqrt{3}$

simplify,

$\sqrt{48} = 4 \cdot \sqrt{3}$

$= 4 \sqrt{3}$

since the rule of radical signs is just like variables we can combine like terms, with a slight difference.

$a \sqrt{b} \pm a \sqrt{b} = \left(a + a\right) \sqrt{b}$

$4 \sqrt{3}$$+$ $\sqrt{3}$

$4 \sqrt{3}$$+$ $\left(1\right)$$\sqrt{3}$

$\textcolor{b l u e}{= 5 \sqrt{3}}$