If each pair of equations $x^2=b_1x+c_1=0,x^2=b_2x+c_2 \text{ and } x^2+b_3x=c_3$ have a common root, prove following

Question

If each pair of equations $x^2=b_1x+c_1=0,x^2=b_2x+c_2 \text{ and } x^2+b_3x=c_3$ have a common root, prove that $(b_1+b_2+b_3)^2=4(c_1+c_2+c_3+b_1b_2)$

My attempt is as follows:

For equations $x^2=b_1x+c_1,x^2=b_2x+c_2$ to have a common root:

$(c_2-c_1)^2=(b_1c_2-b_2c_1)(b_1-b_2)$

For equations $x^2=b_2x+c_2,x^2+b_3x=c_3$ to have a common root:

$(c_3-c_2)^2=(b_2c_3+b_3c_2)(b_3+b_2)$

For equations $x^2=b_1x+c_1,x^2+b_3x=c_3$ to have a common root:

$(c_3-c_1)^2=(b_1c_3+b_3c_1)(b_1+b_3)$

Adding all three equations:

$2({c_1}^2+{c_2}^2+{c_3}^2-c_1c_2-c_2c_3-c_3c_1)=({b_1}^2+b_2b_3)(c_2+c_3)+({b_2}^2+b_1b_3)(c_1+c_3)+({b_3}^2-b_1b_2)(c_1+c_2)$

But from here I was not able to proceed towards the proof. Please help me in this.

Answer 1

I believe the equations are of the form $$x^2+b_1x+c_1=0$$

f $p,q,r$ are the common roots

$2(p+q+r)=-(b_1+b_2+b_3)$

$$\implies2p=b_2+b_3-b_1$$

Again $c_1+c_3=pq+pr=p(q+r)=-pb_2$

$b_2(b_2+b_3-b_1)=-2(c_1+c_3)$

Similarly $b_3(b_3+b_1-b_2)=-2(c_1+c_2)$

and what should be the last?

Add all three

Answer 2

I finally got this, thanks to @lab bhattacharjee

There are three quadratic equations

\begin{equation} x^2-b_1x-c_1=0\tag{1} \end{equation}

\begin{equation} x^2-b_2x-c_2=0\tag{2} \end{equation}

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

Suppose $(1)$ and $(2)$ have a common root as $p$, $(2)$ and $(3)$ have a common root as $q$, (1) and (3) have a common root as $r$.

\begin{equation} p+r=b_1\tag{4} \end{equation}

\begin{equation} p+q=b_2\tag{5} \end{equation}

\begin{equation} -(q+r)=b_3\tag{6} \end{equation}

$$b_1+b_2+b_3=2p$$

\begin{equation} (b_1+b_2+b_3)^2=4p^2\tag{7} \end{equation}

\begin{equation} -pr=c_1\tag{8} \end{equation}

\begin{equation} -pq=c_2\tag{9} \end{equation}

\begin{equation} -qr=c_3\tag{10} \end{equation}

Adding $(8),(9),(10)$

\begin{equation} 4(c1+c2+c3)=-4(pq+qr+pr)\tag{11} \end{equation}

\begin{equation} 4b_1b_2=4(p+r)(p+q)\tag{12} \end{equation}

Adding $(11),(12)$

$$4(c_1+c_2+c_3+b_1b_2)=-4(pq+qr+pr)+4(p^2+pq+pr+qr)$$ $$4(c_1+c_2+c_3+b_1b_2)=4p^2$$

Hence $4(c_1+c_2+c_3+b_1b_2)=(b_1+b_2+b_3)^2$