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Find the condition for the three equations $a_rx^2+b_rx+c_r=0$; $r=1,2,3$ to have a common root.

My attempt is as follows:

\begin{equation} a_1x^2+b_1x+c_1=0\tag{1} \end{equation}

\begin{equation} a_2x^2+b_2x+c_2=0\tag{2} \end{equation}

\begin{equation} a_3x^2+b_3x+c_3=0\tag{3} \end{equation}

Step $1$: Eliminate $c_1,c_2,c_3$

Multiplying $(1)$ with $c_2c_3$, $(2)$ with $c_1c_3$, $(3)$ with $2c_1c_2$.

Adding $(1)$ and $(2)$:

\begin{equation} (a_1c_2c_3+a_2c_1c_3)x^2+(b_1c_2c_3+b_2c_1c_3)x+2c_1c_2c_3=0\tag{4} \end{equation}

Subtracting $(3)$ from $(4)$:

$$(a_1c_2c_3+a_2c_1c_3-2a_3c_1c_2)x^2+(b_1c_2c_3+b_2c_1c_3-2b_3c_1c_2)x=0$$ $$x=\frac{2b_3c_1c_2-b_1c_2c_3-b_2c_1c_3}{a_1c_2c_3+a_2c_1c_3-2a_3c_1c_2}$$ $$x=\frac{c_2(b_3c_1-b_1c_3)+c_1(b_3c_2-b_2b_3)}{c_2(a_1c_3-a_3c_1)+c_1(a_2c_3-a_3c_2)}\tag{5}$$

Step $2$: Eliminate $a_1,a_2,a_3$

Multiplying $(1)$ with $a_2a_3$, $(2)$ with $a_1a_3$, $(3)$ with $2a_1a_2$.

Adding $(1)$ and $(2)$: $$2a_1a_2a_3x^2+(a_2a_3b_1+a_1a_3b_2)x+(a_2a_3c_1+a_1a_3c_2)=0\tag{6}$$

Subtracting $(3)$ from $(6)$:

$$x(a_2a_3b_1+a_1a_3b_2-2a_1a_2b_3)+(a_2a_3c_1+a_1a_3c_2-2a_1a_2c_3)=0$$

$$x=\frac{2a_1a_2c_3-a_2a_3c_1-a_1a_3c_2}{a_2a_3b_1+a_1a_3b_2-2a_1a_2b_3}$$

$$x=\frac{a_2(a_1c_3-a_3c_1)+a_1(a_2c_3-a_3c_2)}{a_2(a_3b_1-a_1b_3)+a_1(a_3b_2-a_2b_3)}\tag{7}$$

Equating $(5)$ and $(7)$, we get

$$\frac{a_2(a_1c_3-a_3c_1)+a_1(a_2c_3-a_3c_2)}{a_2(a_3b_1-a_1b_3)+a_1(a_3b_2-a_2b_3)}=\frac{c_2(b_3c_1-b_1c_3)+c_1(b_3c_2-b_2b_3)}{c_2(a_1c_3-a_3c_1)+c_1(a_2c_3-a_3c_2)}\tag{8}$$

Now equation $8$ is a monster equation, is there any easy way to solve this?

Actual answer is following $$(c_1a_2-c_2a_1)^2=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1) \text{ and } \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}=0$$

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A more direct way.

If those three equations have a common root, it means that the linear combination of the three vectors

$$v_a=\begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \, v_b=\begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}, \, v_c=\begin{pmatrix} c_1\\ c_2\\ c_3 \end{pmatrix}, \,$$

with the coefficients $(x^2, x, 1)$ is equal to zero. Hence those three vectors are linearly dependent and their determinant vanishes.

Bonus question: is the converse true?

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  • $\begingroup$ sorry I didn't understand it, can you please explain in detail , I am just a learner. $\endgroup$ – user3290550 Oct 17 '19 at 14:26
  • $\begingroup$ @user3290550 If you know what a linear combination of vectors is, write the linear combination I suggest. You'll see that having this linear combination equal to zero is equivalent to have $x$ as a root of the three initial equations. $\endgroup$ – mathcounterexamples.net Oct 17 '19 at 14:34
  • $\begingroup$ As you suggested, I wrote $a_1x^2+b_1x+c_1=0\quad$, $a_2x^2+b_2x+c_2=0\quad$, $a_3x^2+b_3x+c_3=0\quad$ It seems x is the common root, and all these three linear equations will be zero simultaneously when x is the common root. But what else we can infer , how to get the required condition. $\endgroup$ – user3290550 Oct 17 '19 at 14:41
  • $\begingroup$ That is what I wrote. The vectors are linearly dependent so their determinant vanishes. $\endgroup$ – mathcounterexamples.net Oct 17 '19 at 14:42
  • $\begingroup$ What is the meaning of that "vectors are linearly dependent", can you please explain. $\endgroup$ – user3290550 Oct 17 '19 at 14:45
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Equivalently you have the system $Av=O$ where $v=(x^2,x,1)^T$ and $A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix}$ and $O=(0,0,0)^T$. For the existence of common roots, you just need that the the system has a solution. As trivial solution (zero solution) is not possible, you have to take $det(A)=0$ for non-trivial solution.

Added-Moreover, removing the third dependent row gives you two equations viz.

$a_1x^2+b_1x+c_1=0$

$a_2x^2+b_2x+c_2=0$

On solving,

$\frac{x^2}{b_1c_2-b_2c_1}=\frac{x}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}$

Or $x=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}$ and $x^2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}$

$\implies (a_2c_1-a_1c_2)^2=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1)$

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  • $\begingroup$ Only $\det(A) =0$ condition is still not sufficient though. It only says that nontrivial of $v$ does exist, but it doesn't tell that the solution is of the form $ v =(v_1, v_2, 1)$ which is required. $\endgroup$ – Azlif Oct 17 '19 at 16:19
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    $\begingroup$ @Azlif..Thanks a lot for pointing out the incompleteness in my answer. $\endgroup$ – Nitin Uniyal Oct 17 '19 at 16:53
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Lemma. For a field $\mathbb{K}$, let $a_i,b_i,c_i\in\mathbb{K}$ for $i\in\{1,2\}$ be such that $a_i\neq 0$ for every $i\in\{1,2\}$. The polynomials $P_1(x):=a_1x^2+b_1x+c_1$ and $P_2(x):=a_2x^2+b_2x+c_2$ have a common root in an extension of $\mathbb{K}$ if and only if $$(a_1c_2-a_2c_1)^2=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1)\,.\tag{$\star$}$$

Let $z$ be a common root of the polynomials $a_1x^2+b_1x+c_1$ and $a_2x^2+b_2x+c_2$. Thus, $$a_1z^2+b_1z+c_1=0\,,\tag{1}$$ and $$a_2z^2+b_2z+c_2=0\,.\tag{2}$$ If $z=0$, then $c_1=c_2=0$. Then, ($\star$) follows immediately. We assume from now on that $z\neq 0$.

From (1) and (2), we have $$(a_1c_2-a_2c_1)z^2+(b_1c_2-b_2c_1)z=c_2(a_1z^2+b_1z+c_1)-c_1(a_2z^2+b_2z+c_2)=0$$ and $$(a_2b_1-a_1b_2)z+(a_2c_1-a_1c_2)=a_2(a_1z^2+b_1+c_1)-a_1(a_2z^2+b_2z+c_2)=0\,.$$ That is, $$(a_1c_2-a_2c_1)z^2=-(b_1c_2-b_2c_1)z\text{ and }(a_1c_2-a_2c_1)=-(a_1b_2-a_2b_1)z\,.$$ Multiplying the two equations above yields $$(a_1c_2-a_2c_1)^2z^2=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1)z^2\,.$$ Because $z\neq 0$, we can divide both sides of the equation above by $z^2$ and obtain ($\star$).

Conversely, suppose that ($\star$) is true. If $a_1c_2=a_2c_1$, then $b_1c_2=b_2c_1$ or $a_1b_2=a_2b_1$. Since $a_1$ and $a_2$ are nonzero, this shows that either $c_1=c_2=0$, or $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are proportional. In either case, the polynomials $a_1x^2+b_1x+c_1$ and $a_2x^2+b_2x+c_2$ have a common root.

We now assume that $a_1c_2\neq a_2c_1$. If $$z:=-\frac{b_1c_2-b_2c_1}{a_1c_2-a_2c_1}\,,$$ then $$(a_1c_2-a_2c_1)z^2+(b_1c_2-b_2c_1)z=0$$ and $$(a_1b_2-a_2b_1)z+(a_1c_2-a_2c_1)=0\,.$$ That is, $$c_2\,P_1(z)-c_1\,P_2(z)=0$$ and $$-a_2\,P_1(z)+a_1\,P_2(z)=0\,.$$ This implies $$(a_1c_2-a_2c_1)\,P_1(z)=a_1\,\big(c_2\,P_1(z)-c_1\,P_2(z)\big)+c_1\,\big(-a_2\,P_1(z)+a_1\,P_2(z)\big)=0\,.$$ As $a_1c_2-a_2c_1\neq 0$, we get $P_1(z)=0$. Similarly, $P_2(z)=0$. Therefore, $z$ is a common root of $P_1(x)$ and $P_2(x)$.

Proposition. For a field $\mathbb{K}$, let $a_i,b_i,c_i\in\mathbb{K}$ for $i\in\{1,2,3\}$ be such that $a_i\neq 0$ for every $i\in\{1,2,3\}$. The polynomials $P_1(x):=a_1x^2+b_1x+c_1$, $P_2(x):=a_2x^2+b_2x+c_2$, and $P_3(x):=a_3x^3+b_3x+c_3$ have a common root in an extension of $\mathbb{K}$ if and only if $$(a_ic_j-a_jc_i)^2=(b_ic_j-b_jc_i)(a_ib_j-a_jb_i)\tag{*}$$ for all $i,j\in\{1,2,3\}$ with $i<j$, and $$\det\left(\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{bmatrix}\right)=0\,.\tag{#}$$

First, let $z$ be a common root of the polynomials $a_1x^2+b_1x+c_1$, $a_2x^2+b_2x+c_2$, and $a_3x^3+b_3x+c_3$. Thus, $$a_1z^2+b_1z+c_1=0\,,$$ $$a_2z^2+b_2z+c_2=0\,,$$ and $$a_3z^2+b_3z+c_3=0\,.$$ Therefore, using an argument by Nitin Uniyal or mathcounterexamples.net, we have (#). The lemma above proves (*).

We shall now prove the converse. Suppose that both (*) and (#) hold. By the converse of the lemma, we know that $P_i(x)$ and $P_j(x)$ has a common root $z_{i,j}$ for every $i,j\in\{1,2,3\}$ such that $i<j$. If there are two different pairs $\{i,j\}$ with the same value $z_{i,j}$, then the three polynomials have a common root, and we are done. We shall prove that this is indeed the case by contradiction. Suppose that $z_{1,2}$, $z_{1,3}$, and $z_{2,3}$ are all distinct.

This implies $$P_1(x)=a_1\,(x-z_{1,2})\,(x-z_{1,3})\,,$$ $$P_2(x)=a_2\,(x-z_{1,2})\,(x-z_{2,3})\,,$$ and $$P_3(x)=a_3\,(x-z_{1,3})\,(x-z_{2,3})\,.$$ Suppose that there exist $\lambda_1,\lambda_2,\lambda_3\in\mathbb{K}$ such that $$\lambda_1\,P_1(x)+\lambda_2\,P_2(x)+\lambda_3\,P_3(x)=0\,.\tag{$\triangle$}$$ Plugging in $x:=z_{1,2}$ in ($\triangle$), we get $\lambda_3=0$. Similarly, $\lambda_1=0$ and $\lambda_2=0$. This means $P_1(x)$, $P_2(x)$, and $P_3(x)$ are linearly independent polynomials, but this contradicts (#). Thus, the proof is now complete.

Remarks.

  1. Note that the assumption that the coefficients of the quadratic terms are nonzero is essential. You can find counterexamples to both the lemma and the proposition if you ignore this assumption. (Nonetheless, the lemma and the proposition still work if we allow some quadratic coefficients to be zero but maintain that at least one quadratic coefficient is nonzero.)
  2. Both the lemma and the proposition do not guarantee that the common root will be in $\mathbb{K}$. Nonetheless, the only way that the common root is not in $\mathbb{K}$ is that there exists a monic polynomial $Q(x)\in\mathbb{K}[x]$ which is irreducible over $\mathbb{K}$ such that $P_i(x)=a_i\,Q(x)$ for every $i$.
  3. In the proposition, the condition (*) or the condition (#) alone is not sufficient for the polynomials to have a common root.
  4. Furthermore, in the proposition, you cannot use only one pair $\displaystyle\{i,j\}\in\binom{\{1,2,3\}}{2}$ to verify (*) along with (#) in the proposition to establish that the polynomials have a common root.
  5. However, in the proposition, it suffices to use only two pairs $\displaystyle\{i,j\}\in\binom{\{1,2,3\}}{2}$ to verify (*) along with (#) to show whether the polynomials have a common root. (The two pairs can be chosen arbitrarily.) I leave the proof of this part to you.
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