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Exactly One Common Root of ax^2-(a+b)x+b and bx^2+(b-c)x-c | JEE

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Question
If the equations
ax2−(a+b)x+b=0andbx2+(b−c)x−c=0ax^{2} - (a+b)x + b = 0 \qquad \text{and} \qquad bx^{2} + (b-c)x - c = 0
 have exactly one root in common (a,b,c≠0)\left(a, b, c \ne 0\right), then which of the following can be correct?
Ab2=acb^{2} = accorrect
B−a=b≠c-a = b \ne ccorrect
Cb=a≠cb = a \ne c
D−a≠b=c-a \ne b = ccorrect
Solution
Step 1: Factorise the first equation instead of using the discriminant - it splits neatly.
ax2−(a+b)x+b=ax2−ax−bx+b=ax(x−1)−b(x−1)=(ax−b)(x−1).ax^{2} - (a+b)x + b = ax^{2} - ax - bx + b = ax(x-1) - b(x-1) = (ax - b)(x-1) .
 So its roots are
x=baandx=1.x = \frac{b}{a} \quad \text{and} \quad x = 1 .
 Step 2: Factorise the second equation the same way.
bx2+(b−c)x−c=bx2+bx−cx−c=bx(x+1)−c(x+1)=(bx−c)(x+1).bx^{2} + (b-c)x - c = bx^{2} + bx - cx - c = bx(x+1) - c(x+1) = (bx - c)(x+1) .
 So its roots are
x=cbandx=−1.x = \frac{c}{b} \quad \text{and} \quad x = -1 .
 Step 3: List the ways one root can be shared. Comparing the two root-pairs {ba, 1}\left\{\frac{b}{a},\, 1\right\} and {cb, −1}\left\{\frac{c}{b},\, -1\right\}, the possibilities are
(i) ba=cb,(ii) ba=−1,(iii) 1=cb.\text{(i)}\ \frac{b}{a} = \frac{c}{b}, \qquad \text{(ii)}\ \frac{b}{a} = -1, \qquad \text{(iii)}\ 1 = \frac{c}{b} .
 (The pairing 1=−11 = -1 is impossible.) Step 4: Case (i). Cross-multiplying,
b2=ac,b^{2} = ac ,
 which is option (1). So (1) can be correct. Step 5: Case (ii). Here b=−ab = -a, and the common root is −1-1. For the sharing to be *exactly* one root we must not also have 1=cb1 = \dfrac{c}{b}, i.e. we need b≠cb \ne c. That is
−a=b≠c,-a = b \ne c ,
 which is option (2). So (2) can be correct. Step 6: Case (iii). Here c=bc = b, and the common root is 11. Again, to avoid a second common root we need ba≠−1\dfrac{b}{a} \ne -1, i.e. b≠−ab \ne -a. That is
−a≠b=c,-a \ne b = c ,
 which is option (4). So (4) can be correct. Step 7: Test option (3), b=a≠cb = a \ne c. Then ba=1\dfrac{b}{a} = 1, so the first equation has the repeated root 11, and the second has roots cb\dfrac{c}{b} and −1-1. A common root would force cb=1\dfrac{c}{b} = 1, i.e. c=b=ac = b = a, contradicting a≠ca \ne c. So there is no common root at all and (3) is not possible. Answer: (1), (2) and (4).
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