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Find the values of k for which the quadratic equation (3k + 1)x² + 2(k+1)x +1= 0 has real and equal roots

Posted by Suman Chakma 3 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Suman Chakma 3 years, 3 months ago

Given equation is y=(3k+1)x 2 +2(k+1)x+1=0 Also, it is given that the equation has equal roots. Then D=0⇒b 2 −4ac=0 a=3k+1 b=2(k+1) c=1 b 2 −4ac=[2(k+1)] 2 −4(3k+1)=0 ⇒4k 2 +4+8k−12k−4=0 ⇒4k 2 −4k=0 ⇒4k(k−1)=0 ⇒k=0 k=1 When k=0 equation y=x 2 +2x+1=0 Roots are x=−1,−1 When k=1

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