If a,b,c are positive and system of equations ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0 has non-trivial solutions. Then the roots of the equation at2+bt+c=0 are
A
real and opposite in sign
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B
both positive
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C
at least one positive
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D
non real
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Solution
The correct option is D non real System of equations has non-trivial solutions ∣∣
∣∣abcbcacab∣∣
∣∣=0&a3+b3+c3−3abc=0
&(a+b+c)(aw2+bw+c)(aw+bw2+c)=0 ∵ a,b,c are positive
∴a+b+c≠0 either aw2+bw+c=0 or aw+bw2+c=0 Hence roots of the equation at2+bt+c=0 are w and w2 i.e. at2+bt+c=0 has non real roots. Ans: D