Question

How would you simplify `sqrt48 + sqrt3`?

Answer 1

`5sqrt3`

Explanation:

We can split up `sqrt48` using the rule that `sqrt(ab)=sqrtasqrtb`.

Thus, `sqrt48=sqrt(16*3)=sqrt16sqrt3=4sqrt3`.

The original expression can be rewritten as

`4sqrt3+sqrt3=sqrt3(4+1)=5sqrt3`

Answer 2

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

`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:

`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(16*3)+sqrt(3)`

`sqrt(16)*sqrt(3)+sqrt(3)`

`4*sqrt(3)+sqrt(3)`

`5sqrt(3)`

Answer 3

`5sqrt3`

Explanation:

radicals in simplified form are `asqrtb`
where a is a rational number.

to begin with , simplify `sqrt48`

by considering the factors of 48 , particularly 'squares'

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

using the following : `sqrta xx sqrtb hArr sqrtab`

`sqrt48 = sqrt16 xx sqrt3 = 4sqrt3`

hence `sqrt48 + sqrt3 = 4sqrt3 + sqrt3 = 5sqrt3`

Answer 4

`color(blue)(5sqrt3)`

:)

Explanation:

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

Since, `sqrt(ab) = sqrta*sqrtb` (Theorem)

`sqrt48 + sqrt 3`

PerfectSquares:

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

we can simplify `sqrt48`, as 48 evenly divides with `16`,

we get,

`sqrt48 = sqrt(16*3)`

using theorem from above,

`sqrt48 = sqrt(16)*sqrt(3)`

simplify,

`sqrt48 = 4*sqrt3`

`=4sqrt3`

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

`asqrtb+-asqrtb=(a+a)sqrtb`

`4sqrt3` `+` `sqrt3`

`4sqrt3` `+` `(1)` `sqrt3`

`color(blue)(= 5sqrt3)`