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Theory of Equations: Value Equations Root Common

JEE Maths question with a full step-by-step solution.

Question
A value of bb for which the equations x2+bx−1=0x^2 + bx - 1 = 0 and x2+x+b=0x^2 + x + b = 0 have one root in common is
A−2-\sqrt{2}
B−i3-i\sqrt{3}correct
Ci5i\sqrt{5}
D2\sqrt{2}
Solution
Step 1: Let rr be the common root. Subtracting the two equations:
(b−1)r−(1+b)=0  ⟹  r=1+bb−1(b-1)r - (1+b) = 0 \implies r = \frac{1+b}{b-1}
 Step 2: Substituting into r2+r+b=0r^2 + r + b = 0 and multiplying through by (b−1)2(b-1)^2:
(1+b)2+(1+b)(b−1)+b(b−1)2=0(1+b)^2 + (1+b)(b-1) + b(b-1)^2 = 0
(1+b)(2b)+b(b−1)2=0  ⟹  b[2+2b+(b−1)2]=0  ⟹  b(b2+3)=0(1+b)(2b) + b(b-1)^2 = 0 \implies b\bigl[2+2b+(b-1)^2\bigr] = 0 \implies b(b^2+3) = 0
 Step 3: b=0b = 0 or b=±i3b = \pm i\sqrt{3}. The required value is b=−i3b = -i\sqrt{3}. Correct answer: (2)
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