Find the value of k such that the quadratic equation has real and equal roots : 4x^2 - 2(k+1) + (k+1).

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As it has equal and real roots b^2-4ac=0 b=-2(k+1) a=4 c=k+1 (-2(k+1))^2 -4(4)(k+1)=0 4(k+1)^2 -16(k+1)=0 4(k+1)[k+1-4]=0 (K+1)(k-3)=0 So k=-1,3
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What are you looking for?