A) : 2 : 3
B) 2 : 3 : 4
C) 4 : 3 : 2
D) 3 : 2 : 1
Correct Answer: B
Solution :
Let \[\alpha ,\beta \]be the common roots of both the equations. For first equation\[a{{x}^{2}}+bx+c=0,\]we have \[\alpha +\beta =\frac{-b}{a}\] ?.(1) \[\alpha .\beta =\frac{c}{a}\] ?..(2) For second equation \[2{{x}^{2}}+3x+4=0,\]we have \[\alpha +\beta =\frac{-3}{2}\] ...(3) \[\alpha .\beta =\frac{2}{1}\] ?(4) Now, from (1) & (3) & from (2) & (4) \[\frac{-b}{a}=\frac{-3}{2}\] \[\frac{c}{a}=\frac{2}{1}\] \[\frac{b}{a}=\frac{3/2}{1}\] Therefore on comparing we get \[a=1,b=\frac{3}{2}\And c=2\]putting these values in first equation, we get\[{{x}^{2}}+\frac{3}{2}x+2=0\]or\[2{{x}^{2}}+3x+4=0\]from this, we get a = 2, b = 3; c = 4 or a : b : c = 2 : 3 : 4
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