Question

The equation $x^2-4x+k=0$ and $x^2+kx-4=0$ , where $k$ is a real number, have exactly one common root .What is the value of $k$?

Answer 1

For common root say $\alpha ,\alpha$ must satisfy both equations

$\Rightarrow {\alpha }^2-4\alpha +k=0$ and

$\Rightarrow {\alpha }^2+kx-4=0$

Elimination of $\alpha$

$\frac{{\alpha }^2}{16-k^2}=\frac{\alpha }{4+k}=\frac{1}{k+4}$

$(4+k{)}^2=(4+k\left)\right(16-k^2)$

$16+k^2+8k=64-4k^2+16k-k^3$

$k^3+5k^2-8k-48=0$

$(k-3)(k^2+8k+16)=0$

$(k-3)(k+4{)}^2=0$

$\Rightarrow k=3,-4$