How do you evaluate #sqrt32 + 2sqrt18 - 3sqrt50 + 3sqrt98#?

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Answer 1 · 2018-03-16T12:44:18

#16sqrt2#

#sqrt16 timessqrt 2 = sqrt32#
#2 times sqrt9 times sqrt2 = 2sqrt18#
#-3 times sqrt25 times sqrt2 = -3sqrt50#
#3 times sqrt49 times sqrt2 = 3sqrt98#

#4sqrt2 = sqrt32#
#6sqrt2 = 2sqrt18#
#-15sqrt2 = -3sqrt50#
#21sqrt2 = 3sqrt98#

#4+6-15+21 = 16sqrt2#
You can combine them because they all have #sqrt2#

Answer 2 · 2018-03-16T12:50:53

See below

#sqrt32+2sqrt18-3sqrt50+3sqrt98=sqrt(8·4)+2sqrt(9·2)-3sqrt(10·5)+3sqrt(49·2)=sqrt(2^2·2·2^2)+2sqrt(3^2·2)-3sqrt(5^2·2)+3sqrt(7^2·2)=#

Due to this obvious identity we can extract some factor from square root simbol (this indentity is #sqrt(a^2)=a#)

#4sqrt2+6sqrt2-15sqrt2+21sqrt2=16sqrt2#

Answer 3 · 2018-03-16T13:00:01

#=16sqrt2#

First write each radicand as the product of its factors, using square numbers wherever possible,

#sqrt32" " +" " 2sqrt18" " -" " 3sqrt50" " +" " 3sqrt98#

#=sqrt(16xx2)" " + 2sqrt(9xx2)" " - 3sqrt(25xx2)" " + 3sqrt(49xx2)#

Find the square roots where possible

#=4sqrt2" " +" " 2xx3sqrt2" "-" " 3xx5sqrt2" + " 3xx7sqrt2#

#=4sqrt2" " +" " 6sqrt2" "-" " 15sqrt2" + " 21sqrt2#

Now you can simplify because they all have #sqrt2#

#=16sqrt2#