# Find the values of k in the equation x^2 + (k+1)x + ((k+1)/4) = 0 if : One root is 1 and when the roots are real.

*print*Print*list*Cite

Expert Answers

justaguide | Certified Educator

The roots of the quadratic equation `x^2 + (k+1)x + ((k+1)/4)` = 0 are given by `(-(k+1)+-sqrt((k+1)^2 - 4*1*(k+1)/4))/2`

= `(-(k+1)+-sqrt((k+1)^2 - (k+1)))/2`

If the roots are real `(k+1)^2 - k - 1 >= 0`

=> `(k+1)(k+1 - 1) >= 0`

=> `k*(k+1) >= 0`

This is true when `k >= 0` and `k+1>= 0` or `k<= 0` and `k+1<= 0`

=> `k>= 0` or `k<= -1`

If one of the roots is equal to 1

`(-(k+1)+-sqrt((k+1)^2 - (k+1)))/2 = 1`

=> `-(k+1)+-sqrt((k+1)^2 - (k+1))) = 2`

=> `(k+1)^2 - (k+1) = (3 + k)^2`

=> `k^2 + 2k + 1 - k - 1 = 9 + 6k + k^2`

=> `k = 9 + 6k`

=> `k = -9/5`

**The roots are real when `k in (-oo, -1]U[0, oo)` . One of the roots is 1 when **`k = -9/5`