<s>[INST] Find k if the two polynomials have one common root The equations $x^2−4x+k=0$ and $x^2+kx−4=0$, where $k$ is a real number, have exactly one common root. What is the value of $k$? 
I know the answer but can it be done with the relation of roots.
$a$ and $b$ are roots of equation 1 and $a$ and $c$ are roots of equation 2. So the relations are :-


*

*$a + b = 4$

*$ab = k$

*$a + c = (-k)$

*$ac= (-4)$


I tried doing all the stuff but couldn't get it. Can we find it using these 4 equations?
 [/INST] If $a+b=4$, then $a^2 + ab = 4a$. So 
$$a^2-4a+k=0. \tag 1$$
If $a+c=-k$, then $a^2+ac=-ka$. So 
$$a^2 + ka -4 = 0. \tag 2$$
So $0 = (a^2 + ka -4)-(a^2-4a+k)=(k+4)a-(k+4)=(k+4)(a-1)$
So $k=-4$ or $a=1$
If $k=-4$, then both trinomials become $x^2-4k-4$ and so they have two roots in common: $2+\sqrt 8$ and $2-\sqrt 8$.
If $a=1$, then plugging $a=1$ into equations $(1)$ and $(2)$ yields the equations
$$1-4+k=0 \qquad \text{and} \qquad 1+k-4=0$$
and both equations have the single solution $k=3$.
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