for what value of 'k' will the following quadratic equation:

(k+1)x2-4kx+9=0 have real and equal roots? Solve the equation.

 

Asked by Ujjwal | 7th Jan, 2018, 03:16: PM

Expert Answer:

begin mathsize 16px style Given space equation space is space left parenthesis straight k plus 1 right parenthesis straight x squared minus 4 kx plus 9 equals 0
Here comma space straight a equals straight k plus 1 comma space straight b equals negative 4 straight k comma space straight c equals 9
For space roots space to space be space real space and space equal comma
Discriminant equals 0
rightwards double arrow straight b squared minus 4 ac equals 0
rightwards double arrow left parenthesis negative 4 straight k right parenthesis squared minus 4 left parenthesis straight k plus 1 right parenthesis left parenthesis 9 right parenthesis equals 0
rightwards double arrow 16 straight k squared minus 36 straight k minus 36 equals 0
rightwards double arrow 4 straight k squared minus 9 straight k minus 9 equals 0
rightwards double arrow 4 straight k squared minus 12 straight k plus 3 straight k minus 9 equals 0
rightwards double arrow 4 straight k left parenthesis straight k minus 3 right parenthesis plus 3 left parenthesis straight k minus 3 right parenthesis equals 0
rightwards double arrow left parenthesis straight k minus 3 right parenthesis left parenthesis 4 straight k plus 3 right parenthesis equals 0
rightwards double arrow straight k equals 3 space or space straight k equals negative 3 over 4 end style

Answered by Rashmi Khot | 8th Jan, 2018, 02:20: PM