# 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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### 1 Answer

`(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 than 0.

Discriminant(`Delta` ) of a quadratic equation ax^2+bx+c = 0 is given by;

`Delta = b^2-4ac`

For our question;

`b = 6`

`a = (k+3)`

`c = (k-5)`

`Delta = 6^2-4xx(k+3)(k-5) `

`Delta > 0`

`6^2-4xx(k+3)(k-5) > 0`

`36-4(k^2-2k-15) > 0`

`36-4k^2+8k+60`

`4(-k^2+2k+24) > 0`

`-k^2+2k+24> 0`

`k^2-2k-24 < 0`

*So the requirement is satisfied.*

`k^2-2k-24 < 0`

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