Question

3root98-root405+root500-root72?

Answer 1

`15sqrt2+sqrt5`

Explanation:

`3sqrt98 - sqrt405+sqrt500-sqrt72`
(I take "3root" to mean 3 times the square root, i.e. `3sqrt98`, and not "cubic root", i.e. `root3(98)`.)

We need to factorise this expression to see if there are any common factors here. We notice that:
`98=2*49=2*7^2`
`405=9*45=5*9^2` (9 is a factor is 405 since 4+0+5=9)
`500=25*20=5^2*5*4=5^3*2^2=5*10^2`
`72=9*8=2*6^2`

Therefore:
`3sqrt(2*7^2) - sqrt(5*9^2)+sqrt(5*10^2)-sqrt(2*6^2)`
Move the square numbers outside and take the square root of each:
`3*7sqrt2-9sqrt5 + 10sqrt5-6sqrt2`
Rearrange so that roots that belong together, come together:
`(21-6)sqrt2+(10-9)sqrt5=15sqrt2+sqrt5`

Answer 2

`15sqrt2+sqrt5`

Explanation:

`3sqrt98-sqrt 405+sqrt 500-sqrt 72`

`:.=3sqrt(2*7*7)-sqrt(5*9*9)+sqrt(5*10*10)-sqrt(3*3*2*2*2)`

`:.=3*7 sqrt 2-9 sqrt 5+10 sqrt 5-2*3 sqrt 2`

`:.=21 sqrt 2-9 sqrt 5+10 sqrt 5-6 sqrt 2`

`:.=15 sqrt 2+sqrt 5`