Find the values of k for which (k + 4)x^2 + (k + 1)x = -1 has equal roots.
+2668 Add 1 to both sides of the equation to make a quadratic: \((k+4)x^2 + (k+1)x+1=0\)
The quadratic has exactly 1 solution when the discriminant(\(b^2 - 4ac \)) equals 0.
This means we have the equation: \((k+1)^2 - 4(k+4) = 0 \)
Simplifying the left-hand side gives: \(k^2−2k−15=0 \), which can be factored into \((k+3)(k−5)=0\), meaning \(k = \color{brown}\boxed{-3, 5}\)
