I'm stuck trying to figure out 8 square of 2 and 13 square root 3 in radical form i know the decimal is 33.83036899 but don't now the radical form can someone please help?

8 #sqrt2# +13 #sqrt3#

8 #sqrt2# +13 #sqrt3#

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Feb 9, 2018

Answer:

#8sqrt(2)+13sqrt(3)# is already in simplest form.

Explanation:

The square roots #sqrt(2)# and #sqrt(3)# are independent over the rationals.

In fact the set of numbers of the form #a+bsqrt(2)+csqrt(3)+dsqrt(6)# where #a, b, c, d# are rational is closed under addition, subtraction, multiplication and division of non-zero numbers, with the coefficients #a, b, c, d# being unique.

For example, #sqrt(3)# is not expressible in the form #a+bsqrt(2)# for rational #a, b#.

Attempting to solve:

#3 = (a+bsqrt(2))^2 = (a^2+2b^2)+2ab sqrt(2)#

We find that we require #a = 0# and/or #b = 0# in order that the right hand side is rational.

If #b=0# then we have #a^2=3#, which is not possible.

If #a=0# then we have #2b^2=3#, so #(2b)^2 = 6#, which is not possible either since #sqrt(6)# is irrational.

Having shown #sqrt(3)# is not expressible in the form #a+bsqrt(2)#, suppose:

#a+bsqrt(2)+csqrt(3)+dsqrt(6) = 0#

Then if at least one of #a, b, c, d# is non-zero, then #c^2-2d^2 != 0# and we find:

#sqrt(3) = -(a+bsqrt(2))/(c+dsqrt(2))#

#color(white)(sqrt(3)) = -((a+bsqrt(2))(c-dsqrt(2)))/(c^2-2d^2)#

#color(white)(sqrt(3)) = (ad-bc)/(c^2-2d^2)+(bc-ad)/(c^2-2d^2)sqrt(2)#

which contrdicts our earleir finding.

Therefore:

#a+bsqrt(2)+csqrt(3)+dsqrt(6) = 0" " => " "a=b=c=d=0#

Hence the representation #a+bsqrt(2)+csqrt(3)+dsqrt(6)# is unique.

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