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Question
Find the values of k for which the quadratic equation (3k + 1) x2 + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.
Solution
The given quadratic equation is (3k+1)x2+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
⇒4k2+8k+4−12k−4=0
⇒4k2−4k=0
⇒k(k−1)=0
⇒k=0 or 1
Thus, the values of k are 0 and 1.
For k = 0,
(3k+1)x2+2(k+1)x+1=0
⇒x2+2x+1=0
⇒(x+1)2=0
⇒x=−1, −1
For k = 1,
(3k+1)x2+2(k+1)x+1=0
⇒4x2+4x+1=0
⇒(2x+1)2=0
⇒x=−1/2,−1/2
Thus, the equal roots are − 1 and −1/2.
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