Not sure if this is a valid method but I was thinking...
Let #t=0#
#g(0)=sqrt0+sqrt1-4#
#=-3# which is negative
Let #t=9#
#g(9)=sqrt9+sqrt10-4#
#=sqrt10-1#
#~~2.16# which is positive
Since we have one negative and one positive value of #g(t)#, we know the function gives a value of 0 at least once. Now, we need to show that there is only one zero value.
#g(t)=sqrtt+sqrt(1+t)-4#
#g(t)=t^(1/2)+(1+t)^(1/2)-4#
#g'(t)=1/2t^(-1/2)+1/2(1+t)^(-1/2)#
Since #g'(t)# will be positive for all positive values of t, (the coefficients are all positive), the function is increasing for all values of x.
This means that it cannot give a value of #g(t)=0# more than once. We know that it does give such a value at some point as we showed earlier. To conclude, we would write.
Since #g'(t)>0" "AAt>=0, g(t)# is increasing for #AAt>=0. #
#:. g(t)=0# for no more than one value of #t#.
Since there are positive and negative values of #g(t)#, #g(t)=0# for only one distinct value of t.
