1 Answer | Add Yours
(x-3)(x+4)/[(x-3)(x+2)}= 1/x. To find non permissible values of x.
x-3 both in numerator and denominator gets cancelled. So we can write the expression:
(x+4)/(x+2) = 1/x. Since the zero in denominator implies division by zero which is not permissible, x+2 = 0 or x=-2 is not permissible. Also there is x in denominator . So x= 0 is not permissible.
Actually 2x-5/x^2+x+5 and 2x-(5/x^2)+x+5 are same.In such a case, x^2 = 0 or x = 0 is not permissible.
If you mea (2x-5)/(x^2+x+5). Then if the denominator x^2+x+5 = 0, then the roots of the equation (if they are real ) are not permissible.
But since the discriminant , (square of coefficient - 4times product od coefficient of x^2 and constant term) = 1^2-4*1*5 = =-19 is negative, there are no real roots and so the expression x^2+x+5 is non zero. So there are no non permissible real values for the given rational expression.
We’ve answered 319,859 questions. We can answer yours, too.Ask a question