(i) √(100x⁴ + 125x⁴) + 7x + 2 = −4x (ii) ∛(64y³) × 2y + 19y + 7² = −3y +…

Concept

These are two disguised quadratic equations. First simplify every surd and power — ∛(a³) = a, √(a²) = |a| and √1600 = 40 — so each equation reduces to the standard form ax² + bx + c = 0. Then find the roots by factorisation (or the quadratic formula x = [−b ± √(b² − 4ac)] / 2a), and finally test each statement by substituting the required value of Z.

Step 1 — Solve equation (ii) and find Z

1. Simplify the surds: ∛(64y³) = 4y, 7² = 49 and √1600 = 40.

2. Substitute: 4y × 2y + 19y + 49 = −3y + 40, i.e. 8y² + 19y + 49 = −3y + 40.

3. Bring every term to one side: 8y² + 22y + 9 = 0.

4. Factorise: (2y + 1)(4y + 9) = 0, so y = −1/2 or y = −9/4. The smallest root is −9/4.

5. Z = smallest root × 4/5 = (−9/4) × (4/5) = −9/5 = −1.8.

Step 2 — Solve equation (i)

1. Simplify the radical: √(100x⁴ + 125x⁴) = √(225x⁴) = 15x².

2. The equation becomes 15x² + 7x + 2 = −4x.

3. Bring every term to one side: 15x² + 11x + 2 = 0.

4. Factorise: (5x + 2)(3x + 1) = 0, so x = −2/5 or x = −1/3. The largest root is −1/3.

Step 3 — Test each statement with Z = −1.8

• (A) Z < −2 → −1.8 < −2 is false, because −1.8 lies to the right of −2 on the number line.

• (B) 2/7 > Z × (−4/27) → Z × (−4/27) = (−9/5)(−4/27) = 4/15. Compare 2/7 and 4/15 by cross-multiplication: 2 × 15 = 30 and 4 × 7 = 28; since 30 > 28, 2/7 > 4/15 is true.

• (C) Z < largest root of equation (i) → −1.8 < −1/3 is true, since −1.8 lies well to the left of −0.33 on the number line.

Cross-check & result

Only (A) fails; both (B) and (C) hold. Quick check on (B): 4/15 ≈ 0.267 and 2/7 ≈ 0.286, confirming 2/7 is the larger. Hence both statement (B) and statement (C) are true.