If the graph of the function f(x)=x^2+2(k-1)x+k+5,k?R cuts the x-axis at least at one point on the positive side, find the set of possible values of the constant “K”

Dear student,

Put Discriminat D>0

D=b2-4ac

for 1 cut on + side--> one root should be positive and one negative

so for this their product is neg. i.e b/c = 2(k-1)/(k+5) < 0 multiplying boyh up n down by (k+5) ( this is done so as to make deno positive as w dont know k-1 is positive or negative

==>  (k+5)(k-1) <0

==>   k belons to  (-5,1)

for tw0 cuts= both roots positive i.e b/c>0 and -b/a>0

==>   2(k-1)/(k+5)  >0   and   -2(k-1) >0

==>    (k+5) k-1 >0   n             k-1<0

         k belongs to (-~,-5) U (1,+~)      n       k    blngs to  (-~,1)

     so, for above commom slution is (-~,-5)

thus final solution is bay taking union of both

k belongs to (-~,-5) U (-5,1)