Question

How do you simplify 4 * the square root of 7 + 8 * the square root of 63?

Answer 1

`28sqrt7`

Explanation:

`"using the "color(blue)"law of radicals"`

`•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)`

`rArrsqrt63=sqrt(9xx7)=sqrt9xxsqrt7=3sqrt7`

`rArr4sqrt7+(8xx3sqrt7)`

`=4sqrt7+24sqrt7`

`=28sqrt7`

Answer 2

`28sqrt7`

Explanation:

Here,

`4sqrt7+8sqrt63=4sqrt7+8sqrt(9xx7`

`=4sqrt7+8*3sqrt7`

`=4sqrt7+24sqrt7`

`=28sqrt7`

Answer 3

`28sqrt(7)`

Explanation:

Assumption: You mean `color(white)("d")4sqrt(7)+8sqrt(63)`

7 is a prime number so lets leave that for the moment.

Consider `sqrt(63)`

Useful trick follows.
note that for 63 `->6+3=9` and as 9 is divisible by 3 then so is 63

`sqrt(63) =sqrt(3xx21) = sqrt(3xx3xx7) larr" The 7 is useful!"`
`color(white)("dd...")=3sqrt(7)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all back together we have:

`4sqrt(7)+(8xx3sqrt(7))`

`4sqrt(7)+24sqrt(7)`

Factor out the `sqrt(7)`

`sqrt(7)(4+24)`

`28sqrt(7)`