SOLUTION: find K so that one zero of the polynomial 2kx^2-20x+21 =0 exceeds the other by 2

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Question 1071073: find K so that one zero of the polynomial 2kx^2-20x+21 =0 exceeds the other by 2

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!


%282%2Asqrt%28400-4%2A2k%2A21%29%29%2F%284k%29=2

%282%2A2%2Asqrt%28100-42k%29%29%2F%284k%29=2

sqrt%28100-42k%29%2Fk=2

square both sides.
%28100-42k%29%2Fk%5E2=4

100-42k=4k%5E2

50-21k=2k%5E2

highlight_green%282k%5E2%2B21k-50=0%29

Resorting to general solution of quadratic equation for this ,
k=%28-21%2B-+sqrt%2821%5E2%2B4%2A2%2A50%29%29%2F4

highlight%28k=%28-21%2B-+29%29%2F4%29-------but I'll leave it to you, to finish this.