Multiple choice

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 $at^2 + bt + c = 0$ are

  1. real and opposite in sign

  2. both positive

  3. at least one positive

  4. non real
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D Correct answer
Explanation

The system has non-trivial solutions if the determinant of the coefficient matrix is zero. This leads to the condition a^3 + b^3 + c^3 - 3abc = 0, which factors as (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0. Since a, b, c are positive, a+b+c is not zero, so a^2+b^2+c^2-ab-bc-ca = 0, implying a=b=c. The equation becomes at^2 + at + a = 0, or t^2 + t + 1 = 0, which has complex roots.

AI explanation

For the system of homogeneous linear equations to have non-trivial solutions, the determinant of its coefficient matrix must be zero, which gives a^3 + b^3 + c^3 - 3abc = 0. This identity factors to (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0. Since a, b, and c are positive real numbers, the first factor cannot be zero, meaning a^2 + b^2 + c^2 - ab - bc - ca = 0. For the quadratic equation at^2 + bt + c = 0, the discriminant is b^2 - 4ac. From our zero identity, we know b^2 = 4ac - 2a^2 - 2c^2, so substituting this yields a discriminant of -2(a^2 + c^2) < 0. Because the discriminant is strictly negative, the roots are non real.

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