Question

(3k+1)x^2+2(k+1)X +k=0. find k​

Answer 1

Answer:

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

equation 4x

2

+4x+1=0

Roots are x=

2

−1

,

2

−1

Step-by-step explanation:

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