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Q.

If a>b>c  and the system of equations ax+by+cz=0, bx+cy+az=0  and cx+ay+bz=0  has a non-trivial solution, then both the roots of the quadratic equation at2+bt+c=0  are

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a

at least one positive root

b

opposite in sign

c

positive

d

imaginary

answer is D.

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Detailed Solution

|abcbcacab|=a3+b3+c3-3abc=a+b+ca2+b2+c2-ab-bc-ca =(a+b+c)(a+bω+cω2)(a+bω2+cω) ⇒     at2+bt+c=0  has both imaginary roots ω, ω2 or both equal roots t=1,1.
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