Find the value of k for which the quadratic equation $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ has equal roots. Also, find the roots.

Open in App

Solution

For the equation to be a perfect square, the determinant should be zero.
Hence,

Now if k=0 then the equation becomes,

Therefore when k=0 then the roots of the equation are -1 and -1.

Now when k=1 then the equation becomes,

Therefpore when k=1 then the roots of the equation are and

Suggest Corrections

Similar questions

(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

k

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

k x^{2} + 1 - 2 (k - 1) x + x^{2} = 0

Join BYJU'S Learning Program