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Question
Find the values of k for which the quadratic equation `(3k+1)x^2+2(k+1)x+1=0` has real and equal roots.
Solution
The given equation is `(3k+1)x^2+2(k+1)x+1=0`
This is of the form `ax^2+bx+c=0` where `a=3k+1, b=2(k+1) and c=1`
∴`D=b^2-4ac`
=`[2(k+1)]^2-4x(3k+1)xx1`
=` 4(k^2++2k+1)-4(3k+1) `
=`4k^2+8k+4-12k-4`
=`4k^2-4k`
The given equation will have real and equal roots if D = 0.
∴ `4k^2-4k=0`
⇒`4k(k-1)=0`
⇒ `4k(k-1)=0`
⇒`k=0 or k-1=0`
⇒`k=0 or k=1`
Hence, 0 and 1are the required values of k.
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