x^2-9÷x-3=-3 at x=3 but for me this =0÷0 which is N.D.

Question

Vikas TU · Answer

Yes it is 0/0 indeterminant form. And thus we need to solve this by factorisation if it is reducing further . Therefore, (x+3) (x-3) / (x -3) = > x+3 at x = 3 the value it gives is = 3 + 3 = 6.

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x^2-9÷x-3=-3 at x=3 but for me this =0÷0 which is N.D.

x^2-9÷x-3=-3 at x=3 but for me this =0÷0 which is N.D.

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x^2-9÷x-3=-3 at x=3 but for me this =0÷0 which is N.D.

x^2-9÷x-3=-3 at x=3 but for me this =0÷0 which is N.D.

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