Roots where one is the square of the other: For x^2 − b x + c = 0, suppose one root is the square of the other. Identify the correct relation between b and c.

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:Let the roots be r and r^2 (order irrelevant). Then the sum and product must match Vieta’s relations for x^2 − bx + c = 0: r + r^2 = b and r^3 = c. Eliminating r between these two relations yields a polynomial identity connecting b and c only. Given Data / Assumptions: Sum: r + r^2 = b Product: r^3 = c r is real (implied by a standard quadratic context), but the identity holds algebraically. Concept / Approach:Express r^2 as b − r and compute r^3 = r * r^2 = r(b − r) = br − r^2. Replace r^2 again by (b − r) to get r^3 in terms of r and b alone. Then substitute c for r^3 and eliminate r via the earlier linear relation to reach a pure b–c identity. A quick numeric check confirms the final formula. Step-by-Step Solution: From r + r^2 = b ⇒ r^2 = b − rr^3 = r(b − r) = br − r^2 = br − (b − r) = r(b + 1) − bBut r^3 = c ⇒ r(b + 1) = b + c ⇒ r = (b + c)/(b + 1)Also c = r^3 = [(b + c)/(b + 1)]^3 ⇒ c(b + 1)^3 = (b + c)^3This identity simplifies to b^3 = 3bc + c^2 + c (after expansion and cancellation). Verification / Alternative check:Test r = 2 ⇒ b = 2 + 4 = 6, c = 8. Then b^3 = 216 and 3bc + c^2 + c = 3*6*8 + 64 + 8 = 216, confirming the relation. Why Other Options Are Wrong: They do not hold for sample values (e.g., r = 2) and conflict with the derived identity from Vieta’s formulas. Common Pitfalls:Trying to solve for r explicitly and then back-substitute with approximate decimals. Work symbolically to maintain exactness. Final Answer:b^3 = 3bc + c^2 + c

Roots where one is the square of the other: For x^2 − b x + c = 0, suppose one root is the square of the other. Identify the correct relation between b and c.

More Questions from Quadratic Equation

Discussion & Comments