(i) Find the values of k for which the quadratic equation $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ has real and equal roots.
(ii) Find the value of k for which the equations $x^{2} + k (2 x + k - 1) + 2 = 0$ has real and equal roots.

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Solution

(i) $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0 h e r e, a = 3 k + 1 a n d b = 2 [k + 1] = 2 k + 2 a n d c = 1$

roots are equal and real and so discriminant will be zero
$D = b^{2} - 4 a c = 0 \Rightarrow [2 k + 2]^{2} - 4 [3 k + 1] [1] = 0 \Rightarrow 4 k^{2} + 8 k + 4 - 12 k - 4 = 0 \Rightarrow 4 k^{2} - 4 k = 0 \Rightarrow 4 k [k - 1] = 0 \Rightarrow k = 0 a n d 1$

(ii) $x^{2} + k (2 x + k - 1) + 2 = 0 x^{2} + 2 k x + k^{2} - k + 2 = 0 x^{2} + 2 k x + (k^{2} - k + 2) = 0 a = 1 a n d b = 2 k a n d c = k^{2} - k + 2$

Roots are equal and real and so discriminant will be zero

$D = b^{2} - 4 a c = 0 \Rightarrow (2 k)^{2} - 4 (1) (k^{2} - k + 2) = 0 \Rightarrow 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k - 8 = 0 \Rightarrow 4 k = 8 \Rightarrow k = 2$

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