Find three values of $k$ , the quadratic equation $x^{2} - 4 x + k = 0$ will have distinct real roots?

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Solution

$x^{2} - 4 x + k = 0$

$k = 0 \to x^{2} - 4 x = 0 \Rightarrow x (x - 4) = 0 \Rightarrow x = 0, 4$

$k = 3 \to x^{2} - 4 x + 3 = 0 \Rightarrow (x - 1) (x - 3) = 0 \Rightarrow x = 1, 3$

$k = 4 \to x^{2} - 4 x + 4 = 0 \Rightarrow (x - 2)^{2} = 0 \Rightarrow x = \pm 2$

$k = 0, 3, 4, . . . . . . .$

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