If the cubic equation a³ − 4a² + 5a − 2 = 0 has the root x, y, z then find the value of x + y + z.

Correct — D. Vieta's rule: in a³ + pa² + qa + r = 0, the roots add to −p.

Read p, the a² coefficient: p = −4
Sum of roots: x + y + z = −p = −(−4) = 4

Check by factoring: put a = 1, and 1 − 4 + 5 − 2 = 0
Divide out (a − 1): a³ − 4a² + 5a − 2 = (a − 1)²(a − 2)
The roots 1, 1 and 2 add to 4 → option (d)

Why the others are wrong

• (a)
  5 — 5 is the coefficient of a, which Vieta's rule ties to xy + yz + zx = 1 + 2 + 2, the sum of products in pairs, not the sum of roots.
• (b)
  6 — 6 is x² + y² + z² = 1 + 1 + 4, the sum of the squares of the roots. It also equals 4² − 2 × 5.
• (c)
  7 — 7 would need an a² coefficient of −7. The roots here are 1, 1 and 2, and they add to 4.

For a cubic with leading coefficient 1, a³ + pa² + qa + r = 0, with roots x, y, z, Vieta's formulas read:

x + y + z = −p
xy + yz + zx = q
xyz = −r

Here p = −4, q = 5, r = −2, so the sum is 4, the pair-sum is 5 and the product is 2. The roots 1, 1 and 2 confirm all three.

Key facts

• For a³ + pa² + qa + r = 0, the sum of the roots is −p.
• For the same cubic, xy + yz + zx = q and xyz = −r.
• x² + y² + z² = (x + y + z)² − 2(xy + yz + zx).
• If the coefficients add to 0, then a = 1 is a root.

Common traps

• Forgetting the minus sign in −p: the coefficient is −4, so the sum is +4.
• Reading the coefficient of a, 5, as the sum of roots, when it is the pair-sum.

No directly related past PYQ was found.