What is x if #6=7/x+x#?

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Answer 1 · 2018-05-03T04:01:53

Below

#6=7/x+x# where #x!=0#

#7/x=6-x#

#x^2*7/x=x^2(6-x)#

#7x=6x^2-x^3#

#x^3-6x^2+7x=0#

#x(x^2-6x+7)=0#

#x=0# or #x^2-6x+7=0#

For #x^2-6x+7=0#, we need to use the quadratic formula

ie #x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(6+-sqrt(36-28))/(2)#

#x=(6+-2sqrt2)/2#

#x=3+-sqrt2#

BUT looking at #x=0#, it cannot be a solution because of #7/0#

Therefore, the answer is #x=3+-sqrt2#