If #(36+16(5)^(1/2))^(1/2)= a+b(c)^(1/2)# where #a#, #b#, and #c# are positive integers, find #abc#?

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Answer 1 · 2018-04-06T06:49:52

#(36+16(5)^(1/2))^(1/2)= a+b(c)^(1/2)#

#=>(16+20+16(5)^(1/2))^(1/2)= a+b(c)^(1/2)#

#=>(4^2+(2*5^(1/2))^2+2*4*2(5)^(1/2))^(1/2)= a+b(c)^(1/2)#

#=>[(4+2*5^(1/2))^2]^(1/2)= a+b(c)^(1/2)#

#=>4+2*5^(1/2)= a+b(c)^(1/2)#

Comparing both sides we get

#a=4,b=2and c=5#

If #(36+16(5)^(1/2))^(1/2)= a+b(c)^(1/2)# where #a#, #b#, and #c# are positive integers, find #a*b*c#?