Theory of EquationsmediumFree

Theory of Equations: Equations Common Root Value Equal

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

Question
If c2=4dc^2 = 4d and the two equations x2−ax+b=0x^2 - ax + b = 0 and x2−cx+d=0x^2 - cx + d = 0 have one common root, then the value of 2(b+d)2(b+d) is equal to
Aac\dfrac{a}{c}
Bacaccorrect
C2ac2ac
Da+ca + c
Solution
Step 1: c2=4dc^2 = 4d means the discriminant of x2−cx+d=0x^2 - cx + d = 0 is c2−4d=0c^2 - 4d = 0, so this equation has a repeated root c2\dfrac{c}{2}. Step 2: c2\dfrac{c}{2} is also a root of x2−ax+b=0x^2 - ax + b = 0:
c24−a⋅c2+b=0  ⟹  d−ac2+b=0  ⟹  b+d=ac2\frac{c^2}{4} - a\cdot\frac{c}{2} + b = 0 \implies d - \frac{ac}{2} + b = 0 \implies b + d = \frac{ac}{2}
 Step 3: 2(b+d)=ac2(b + d) = ac. Correct answer: (2)
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