The equations $x^{2} - 4x + k = 0$ and $x^{2} + kx - 4 = 0$, where $k$ is a real number, have exactl

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

Accepted Answer

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)$$\begin{matrix} \Rightarrow & & \ \ \ \ \ \ \ \ & \ (k + 4)\alpha - (4 + k) & \ = 0 \ \Rightarrow & & \ \ \ \ \ \ \ \ & \alpha & \ = 1	ext{~or~}k = - 4 \ & & \ \ \ \ \ \ \ \ & \alpha & \ = 1 \Rightarrow k = 3 \end{matrix}$$$k = - 4$ gives both roots common$k = 3$ satisfies the condition.

Question

Options

Solution

More Quadratic Equation Questions