How would you simplify $\sqrt{48} + \sqrt{3}$ $sqrt48 + sqrt3$ ?

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How would you simplify $\sqrt{48} + \sqrt{3}$ $sqrt48 + sqrt3$ ?

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Answer 1 · 2016-02-13T18:54:24

$5 \sqrt{3}$ $5sqrt3$

Explanation:

We can split up $\sqrt{48}$ $sqrt48$ using the rule that $\sqrt{a b} = \sqrt{a} \sqrt{b}$ $sqrt(ab)=sqrtasqrtb$ .

Thus, $\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \sqrt{3} = 4 \sqrt{3}$ $sqrt48=sqrt(16*3)=sqrt16sqrt3=4sqrt3$ .

The original expression can be rewritten as

$4 \sqrt{3} + \sqrt{3} = \sqrt{3} (4 + 1) = 5 \sqrt{3}$ $4sqrt3+sqrt3=sqrt3(4+1)=5sqrt3$

Answer 2 · 2016-02-13T18:56:53

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

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

Explanation:

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

$\frac{48}{3} = 16 \Leftrightarrow 48 = 3 \cdot 16$ $48/3=16<=>48=3*16$

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

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

$\sqrt{16 \cdot 3} + \sqrt{3}$ $sqrt(16*3)+sqrt(3)$

$\sqrt{16} \cdot \sqrt{3} + \sqrt{3}$ $sqrt(16)*sqrt(3)+sqrt(3)$

$4 \cdot \sqrt{3} + \sqrt{3}$ $4*sqrt(3)+sqrt(3)$

$5 \sqrt{3}$ $5sqrt(3)$

Answer 3 · 2016-02-13T19:00:11

$5 \sqrt{3}$ $5sqrt3$

Explanation:

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

to begin with , simplify $\sqrt{48}$ $sqrt48$

by considering the factors of 48 , particularly 'squares'

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

using the following : $\sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a} b$ $sqrta xx sqrtb hArr sqrtab$

$\sqrt{48} = \sqrt{16} \times \sqrt{3} = 4 \sqrt{3}$ $sqrt48 = sqrt16 xx sqrt3 = 4sqrt3$

hence $\sqrt{48} + \sqrt{3} = 4 \sqrt{3} + \sqrt{3} = 5 \sqrt{3}$ $sqrt48 + sqrt3 = 4sqrt3 + sqrt3 = 5sqrt3$

Answer 4 · 2016-02-13T19:05:32

$5 \sqrt{3}$ $color(blue)(5sqrt3)$

:)

Explanation:

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

Since, $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$ $sqrt(ab) = sqrta*sqrtb$ (Theorem)

$\sqrt{48} + \sqrt{3}$ $sqrt48 + sqrt 3$

PerfectSquares:

$4$ $color(blue)(4)$ $= 2 \cdot 2$ $= 2*2$
$9$ $color(blue)(9)$ $= 3 \cdot 3$ $= 3*3$
$16$ $color(blue)(16)$ $= 4 \cdot 4$ $= 4*4$
$25$ $color(blue)(25)$ $= 5 \cdot 5$ $= 5*5$
$36$ $color(blue)(36)$ $= 6 \cdot 6$ $= 6*6$
$49$ $color(blue)(49)$ $= 7 \cdot 7$ $= 7*7$
$64$ $color(blue)(64)$ $= 8 \cdot 8$ $= 8*8$
$81$ $color(blue)(81)$ $= 9 \cdot 9$ $= 9*9$
$100$ $color(blue)(100)$ $= 10 \cdot 10$ $= 10*10$

we can simplify $\sqrt{48}$ $sqrt48$ , as 48 evenly divides with $16$ $16$ ,

we get,

$\sqrt{48} = \sqrt{16 \cdot 3}$ $sqrt48 = sqrt(16*3)$

using theorem from above,

$\sqrt{48} = \sqrt{16} \cdot \sqrt{3}$ $sqrt48 = sqrt(16)*sqrt(3)$

simplify,

$\sqrt{48} = 4 \cdot \sqrt{3}$ $sqrt48 = 4*sqrt3$

$= 4 \sqrt{3}$ $=4sqrt3$

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

$a \sqrt{b} \pm a \sqrt{b} = (a + a) \sqrt{b}$ $asqrtb+-asqrtb=(a+a)sqrtb$

$4 \sqrt{3}$ $4sqrt3$ $+$ $+$ $\sqrt{3}$ $sqrt3$

$4 \sqrt{3}$ $4sqrt3$ $+$ $+$ $(1)$ $(1)$ $\sqrt{3}$ $sqrt3$

$= 5 \sqrt{3}$ $color(blue)(= 5sqrt3)$

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