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Question

The origin and the roots of the equation x2+ax+b = 0 form an equilateral triangle, if :


A

a2=b

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B

b2=3a

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C

b2=a

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D

a2=3b

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Solution

The correct option is D

a2=3b


x2+ax+b = 0 a±a24b2

If a24b0 then the three points, 0 and the two roots are real and so collinear. Therefore, considering a24b<0, we have roots as a±4ba22

Let A = a2+i4ba22, B = a2i4ba22

OA = a24+4ba24 = b = OB

AB = (4ba2). Since OA = OB = AB,

we get 4ba2 = b i.e., a2 = 3b


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