Let equation and (a≠c) have the same root . Please determine this root value using a, b, c, d to represent it.

we can add the two equations to obtain

$2x^2 +(a+c)x+(b+d)=0\\ x^2 + \dfrac{a+c}{2}x+\dfrac{b+d}{2}=0$

$\text{Now just apply the quadratic formula}\\ r_{1,2} = \dfrac{-\frac{a+c}{2}\pm \sqrt{\left(\frac{a+c}{2}\right)^2 - 2(b+d)}}{2}$

$\text{we can clean this up a bit}\\ r_{1,2} = \frac{1}{4} \left(\pm\sqrt{(a+c)^2-8 (b+d)}-a-c\right)$

$\text{If there is only a single shared root }r_1 = r_2\\ \text{if there are no shared roots}\\ (a+c)^2 - 8(b+d)<0$

Why add the two equations ?

Why not simply subtract one from the other ?

Suppose that the first equation has roots p and q and that the second equation has roots p and s,

so that the common root is p.

Then we have

$\displaystyle (x-p)(x-q)=0 \text{ and }(x-p)(x-s)=0\; .$

Subtracting the second equation from the first,

$\displaystyle (x-p)\{(x-q)-(x-s)\}=0\; ,$

so

$\displaystyle (x-p)(s-q)=0\; ,$

showing that  x = p  is the  root of the resulting equation,  $\displaystyle (\;s \neq q\;)\;.$

So, going back to the original equations and subtracting one from the other,

$\displaystyle x(a-c)+(b-d)=0\;,$

$\displaystyle x=\frac{d-b}{a-c}\; .$

Tiggsy