# Find the value of k for which the quadratic equation (k + 4) x^2 +(k +1)x +1 = 0 has equal roots .

Solution: Let D be the discriminant of the given equation.

We have ,        $\\ \Rightarrow a=k+4 ,b=k+1 ;and ;c=1$

$\\\therefore D= b^2-4ac\\ \\\Rightarrow D=(k+1)^2-4(k+4)\\ \\\Rightarrow D=k^2 -2k -15 =(k-5)(k+3)\\ \\\Rightarrow k=5 ,-3$

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