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

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#### Solution

The given quadratic equation is (3k+1)x^{2}+2(k+1)x+1=0.

Since the given quadratic equation has equal roots, its discriminant should be zero.

∴ D = 0

⇒ 4(k+1)^{2}−4×(3k+1)×1=0

⇒4k^{2}+8k+4−12k−4=0

⇒4k^{2}−4k=0

⇒k(k−1)=0

⇒k=0 or 1

Thus, the values of k are 0 and 1.

For k = 0,

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

⇒x2+2x+1=0

⇒(x+1)2=0

⇒x=−1, −1

For k = 1,

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

⇒4x^{2}+4x+1=0

⇒(2x+1)^{2}=0

⇒x=−1/2,−1/2

Thus, the equal roots are − 1 and −1/2.

Concept: Nature of Roots

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