ANother simple one..

1st method: simple one.

subtract the equations. it becomes : x(1-a) + (a-1) = 0

so, x=1 is the common root.

now,putting x=1 we get a=(-2).

2nd method: using the formula :

( a₀b₂-a₂b₀ )² = (a_ob₁-a₁b₀) (a₁b₂-a₂b₁)

if two equations are : a₂x² + a₁x + a₀ and b₂x² + b₁x + b₀

Consider the two quadratic equations
a m² + b m + c = 0
a'm² + b'm + c'= 0
with a and a' not zero, together with 3 determinants

|b c| |a c| |a b|
A = | | B = - | | C = | |
|b' c'| |a' c'| |a' b'|

The equations have a common root if and only if
A.C = B².
The expression A.C = B² is called the eliminant of Sylvester.
Proof:
We multiply the first equation with a', the second one with a, and we calculate the difference of these two results.
(ba' - a'b) m + (a'c - ac') = 0.
From this, the system of the two quadratic equations is equivalent with

/ (ba' - a'b) m + (a'c - ac') = 0.
\ a m² + b m + c = 0

These equations have a common root if and only if the root of the first one, is a root of the second equation. The root of the first equation is
(a'c - a c') / (b'a - a b') = B/C .
The condition then is:

a (B/C)² + b (B/C) + c = 0
<=>
a B² + b B C + c C2 = 0
<=>
a B² = -C ( b B + c C)
<=>
...
<=>
a B² = a ( C A) and since a is not 0
<=>
A.C = B²

this post may be out of topic but still a numerical data, showing the application of common roots concept in curves

Take the following equations with parameter m.

(x - 3)2 + y2 = (5 - m)2

(x + 3)2 + y2 = (5 + m)2

For each value of m we have two fixed circles, the associated curves. When m is a variable, the intersection points of the two circles, form a locus c.
If we consider m as unknown in previous system, the system takes the form

a m2 + b m + c = 0

a'm2 + b'm + c'= 0

The coefficients are functions of x and y.
P(x,y) is a point of the locus if and only if last system has a solution for m. In other words if and only if the two quadratic equations have a common root.

sir,cn this hv a specific graphical proof ?