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.  

Answer 1

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.