How do you simplify #( sqrt 2 + sqrt 5)^4#?

1 Answer
George C.
Apr 11, 2016

#(sqrt(2)+sqrt(5))^4 = 89+28sqrt(10)#

Explanation:

Note that #(sqrt(2)+sqrt(5))^4 = ((sqrt(2)+sqrt(5))^2)^2#

So let us square #(sqrt(2)+sqrt(5))# twice:

#(sqrt(2)+sqrt(5))^2#

#=(sqrt(2))^2+2(sqrt(2))(sqrt(5))+(sqrt(5))^2#

#=2+2sqrt(10)+5#

#=7+2sqrt(10)#

Then:

#(7+2sqrt(10))^2#

#=7^2+2(7)(2sqrt(10))+(2sqrt(10))^2#

#=49+28sqrt(10)+40#

#=89+28sqrt(10)#

Check

Let's check the calculation a different way.

From the Binomial Theorem:

#(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4#

So:

#(sqrt(2)+sqrt(5))^4#

#=(sqrt(2))^4+4(sqrt(2))^3(sqrt(5))+6(sqrt(2))^2(sqrt(5))^2+4(sqrt(2))(sqrt(5))^3+(sqrt(5))^4#

#=4+8sqrt(10)+60+20sqrt(10)+25#

#=89+28sqrt(10)#

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