How do you simplify #sqrt13+sqrt52#?

Question

2214 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T12:06:05

#3sqrt(13)#

Given: #sqrt(13)+sqrt(52)#

We have:

#sqrt(52)#

#=sqrt(13*4)#

Using #sqrt(ab)=sqrt(a)sqrt(b) \ "for" \ a,b>=0#.

#=sqrt(4)*sqrt(13)#

#=2sqrt(13)# (only use principal square root)

So, the expression becomes

#<=>sqrt(13)+2sqrt(13)#

Factoring,

#sqrt(13)(1+2)#

#=3sqrt(13)#

Answer 2 · 2018-03-18T12:13:23

#3sqrt13#

With any surd, we can use the law:

#sqrta xx sqrtb=sqrtab#

Simplifying #sqrt13#

#-># Since it is a prime number it cannot be simplified any further

Simplifying #sqrt52#

We can figure out that we can get #52# from #13 xx4#

#therefore# #-> sqrt13 xxsqrt4#

As said earlier, #sqrt13# cannot be simplified, but #sqrt4# can, using our squared numbers of #1,4,9,12,25...#, #sqrt4=2#

Therefore this turns to:

#2sqrt13# as we always put the value in front of the square root#

Adding the surds:

Since we simplified the surds, we can plug these values back in to get:

#sqrt13+2sqrt13#

Since #sqrt13# means #1sqrt13# we can just add both #1# and #2# to get us #3sqrt13# as the root of #sqrt13# is the same in both expressions.

#therefore# #sqrt13+2sqrt13 -> 3sqrt13#