The equation(k+3)x^2+6x+k = 5, where k is a constant, has two distinct real solutions for x.Show that k satisfies k^2-2k-24<0

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`(k+3)x^2+6x+k = 5`

`(k+3)x^2+6x+k-5 = 0`

 

`(k+3)x^2+6x+k-5 = 0` Since the equation has two distinct real solutions discriminant(`Delta` ) of the equation is greater...

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