Question #f9672

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Answer 1 · 2018-01-04T22:58:56

#k=3#

#(x^2)^n*(k/x)^(5-n)*C(5,n)=270x#

#x^(2n)*k^(5-n)/x^(5-n)*C(5,n)=270x#

#x^(2n-(5-n))*k^(5-n)*C(5,n)=270x#

#x^(3n-5)*k^(5-n)*C(5,n)=270x#

After equating powers of #x#, I found

#3n-5=1#, so #n=2#

After putting #n=2# and simplifications,

#k^(5-2)*C(5,2)=270#

#10k^3=270#

#k^3=20#

Thus, #k=3#