### 1 Answer | Add Yours

Given the quadratic equation :

kx^2 + kx - 3x -3 = 0

==> kx^2 + (k-3) x - 3 = 0

Given that the roots are equal.

==> x1= x2= x.

We know that:

x1+x2= -b/a

==> 2x = (3-k)/k

==> 2x = 3/k - 1...........(1)

x1*x2= c/a

==> x^2 = -3/k ...........(2)

==> 2x = -x^2 -1

==> x^2 +2x +1 = 0

==> (x+1)^2 = 0

==> x = -1

==> k= -3/x^2 = -3

**==> k= -3**

Lets check ....

==> -3x^2 -6x -3 = 0

==> -3(x^2 +2x +1) = 0

==> -3(x+1)^2 = 0

==> x = -1

**Then, the root is -1.**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes