Quadratic EquationHard
Question
The equations $x^{2} - 4x + k = 0$ and $x^{2} + kx - 4 = 0$, where $k$ is a real number, have exactly one common root. Then the number of values of k is
Options
A.0
B.1
C.2
D.3
Solution
Let $\alpha$ be the common root
$$\Rightarrow \ \begin{array}{r} \alpha^{2} - 4\alpha + k = 0\#(1) \\ \alpha^{2} + k\alpha - 4 = 0\#(2) \end{array}$$
(2) and (1)
$$\begin{matrix} \Rightarrow & & \ \ \ \ \ \ \ \ & \ (k + 4)\alpha - (4 + k) & \ = 0 \\ \Rightarrow & & \ \ \ \ \ \ \ \ & \alpha & \ = 1\text{~or~}k = - 4 \\ & & \ \ \ \ \ \ \ \ & \alpha & \ = 1 \Rightarrow k = 3 \end{matrix}$$
$k = - 4$ gives both roots common
$k = 3$ satisfies the condition.
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