Theory of Equations

4

Step 1: Write the coefficients of P(x) in order: +1, −3, +1, +1, −2, +1. Step 2: Count sign changes: +to−(1), −to+(2), +to+(no change), +to−(3), −to+(4). So s₁ = 4. Step 3: By Descartes' Rule, the number of positive real roots cannot exceed s₁ = 4. Step 4: Also, the actual number of positive roots equals 4, 2, or 0 (must differ from s₁ by an even number). Step 5: Therefore the maximum number of positive real roots is 4.

4

Step 1: The polynomial is 2x⁴ + 5x³ − 7x² + 0·x + 8 = 0. Step 2: By Vieta's formula for a degree-4 polynomial ax⁴ + bx³ + cx² + dx + e = 0, the product of all roots = e/a (with sign (−1)⁴ = +1). Step 3: Here a = 2 and the constant term e = 8. Step 4: So αβγδ = (−1)⁴ · (e/a) = 8/2 = 4. Step 5: The formula is: product of roots = constant term / leading coefficient × (−1)ⁿ. For n = 4 (even), the sign is positive. Answer = 4.

ac³ = b³d

Step 1: Let the roots in GP be α/λ, α, αλ. Step 2: Sum: α(1/λ + 1 + λ) = −b/a ... (i). Sum of products two at a time: α²(1/λ + λ + 1) = c/a ... (ii). Product: α³ = −d/a ... (iii). Step 3: Dividing (ii) by (i): α = c/a ÷ (−b/a) = −c/b. Step 4: From (iii): α³ = −d/a. Substituting α = −c/b: (−c/b)³ = −d/a → −c³/b³ = −d/a → ac³ = db³. Step 5: So the required condition is ac³ = b³d.

−1

Step 1: Check whether the equation is Type I or Type II reciprocal equation. Step 2: Coefficients from beginning: 7, −43, −43, 7. Coefficients from end: 7, −43, −43, 7. They are equal, so this is Type I. Step 3: The degree is 3 (odd). For an odd-degree Type I reciprocal equation, x = −1 is always a root. Step 4: Verify: P(−1) = 7(−1) − 43(1) − 43(−1) + 7 = −7 − 43 + 43 + 7 = 0. ✓ Step 5: After dividing by (x+1), we get the quadratic 7x² − 50x + 7 = 0, giving roots 7 and 1/7. So all three roots are −1, 7, 1/7.