<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] Using your equations . . .


*

*$\;a + b = 4$

*$\;ab = k$

*$\;a + c = (-k)$

*$\;ac= (-4)$


Adding equations $(1)$ and $(4)$, we get
$$a + b + ac=0\tag{5}$$
Adding equations $(2)$ and $(3)$, we get
$$a + c + ab=0\tag{6}$$
Subtracting equation $(6)$ from equation $(5)$, we get
\begin{align*}
&b-c + ac - ab = 0\\[4pt]
\implies\;&b-c+a(c-b) = 0\\[4pt]
\implies\;&(b-c)(1-a)= 0\\[4pt]
\implies\;&b=c\;\;\text{or}\;\;a=1\\[4pt]
\end{align*}

If $b=c$, then the left-hand-sides of equations $(2)$ and $(4)$ are equal, hence the right-hand-sides must also be equal, which yields $k=-4$. But for $k=-4$, the two given equations are identical, hence have two common roots, not one.

Hence we must have $a=1$. 

Using $a=1$ in equation $(1)$ yields $b=3$, and then equation $(2)$ yields $k=3$.
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