If x2+x+a=0
and x2+ax+1 have a common root,
find the value of a....
(Give as many proof as possible)
11st 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 = (aob1-a1b0) (a1b2-a2b1)
if two equations are : a2x2 + a1x + a0 and b2x2 + b1x + b0
1Consider 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
1this 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.
341I guess nishant sir posted this one to see if you would miss a = 1 (unless he meant exactly one common root)
1yes ,sir .i missed dat out as on putting a=1, the equation won't carry any real roots.
but,cn't say if he also meant dat. thnx sir fr mentioning it.
62No prophet sir, not exactly that.. that was an observation..
I have a proof which i think is very good.. but is by graphs.. '
so was looking for that one :)
62First observe that the coefficient of x^2 is equal in both the equations x^2+ax+1 and x^2+x+a
a denotes how steep the graph is... because both of these have the same steepness, they will meet only once...
More over this point where they meet is the common root....
also, by observation see that f(1) = g(1)
Hence also, f(1)=0... thus, a=-2