<p>If all the three roots of \(az^3 + bz^2 + cz + d = 0\) have negative real parts (\(a, b, c \in R\)), then</p>
Step-by-Step Solution
Key Concept: Use Routh-Hurwitz stability criterion: for a cubic with real coefficients, all roots have negative real parts iff the polynomial coefficients satisfy specific sign and determinant conditions. For az³+bz²+cz+d=0, we need a,b,c,d all same sign and bc>ad.
<p><strong>Step 1: Apply Routh-Hurwitz for cubic</strong></p><p>For cubic az³+bz²+cz+d=0 with a≠0, all roots have negative real parts iff:</p><ul><li>Condition 1: a, b, c, d all have the same sign</li><li>Condition 2: bc > ad (determinant condition from Routh table)</li></ul><p><strong>Step 2: Establish sign constraints</strong></p><p>If a>0, then b>0, c>0, d>0 (or all negative, equivalent by dividing by -1)</p><p><strong>Step 3: Verify with Vieta's formulas</strong></p><p>If roots are r₁, r₂, r₃ with Re(rᵢ)<0:</p><ul><li>r₁+r₂+r₃ = -b/a < 0 ✓</li><li>r₁r₂+r₂r₃+r₃r₁ = c/a > 0 (sum of products has Re>0)</li><li>r₁r₂r₃ = -d/a < 0 ✓</li></ul><p><strong>Step 4: Conclude conditions</strong></p><p>∴ Answer: <strong>a,b,c,d have same sign AND bc>ad</strong> (Specific options A,B,C,D depend on question choices—typically asking which statements are true)</p>
Correct Answer: A, B, C, D