For what values of z will the following equation have equal roots ?

(z+4) $$\text{x}^2$$​+(z+1)x+1=0 

The quadratic equation is:

For a quadratic equation $$ax^2 + bx + c = 0$$​ to have equal roots, the discriminant must be zero.
The discriminant of a quadratic equation is given by:

​$$D = b^2 - 4ac$$​​

If D = 0, the roots are equal.

For the quadratic equation:

​$$x^2 + (z+1)x + 1 = 0$$​​

a = z + 4

b = z + 1

c = 1

So,

​$$D = (z+1)^2 - 4(z+4)$$​​

For equal roots D = 0

Now,

​$$(z+1)^2 - 4(z+4) = 0$$​​

​$$(z^2 + 2z + 1) - 4z - 16 = 0$$​​

​$$z^2 + 2z + 1 - 4z - 16 = 0$$​​

​$$z^2 - 2z - 15 = 0$$​​

​$$z^2 - 2z - 15 = 0$$​​

​$$z^2 - 2z - 15 = 0$$​​

(z-5)(z+3) = 0

z – 5 = 0 or z + 3 = 0

z = 5 or z = −3

Thus, for z = 5, -3, expression have equal roots.