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Question

If a>b>c and a3+b3+c3=3abc, then the quadratic equations ax2+bx+c=0 has roots which are

A
both real
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B
both +ve
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C
both ve
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D
one +ve and one ve
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Solution

The correct option is A both real
a3+b3+c32abc=12(a+b+c)[(ab)2+(bc)2+(ca)2]=0
[a>b>c, so(ab)2+(bc)2+(ac)2>0]
a+b+c=0 ...(i)
Comparing (i) with ax2+bx+c=0, we find x=1.
Also, ax2+bx+c=0 for xca
Therefore, the roots of ax2+bx+c=0 are 1 and ca.

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