If one root of 7y^2 − 50y + k = 0 is y = 7, find the value of k.

Introduction / Context:
A known root of a polynomial must zero the polynomial when substituted. Use this to solve for the unknown parameter k quickly by plugging y = 7 into the given quadratic expression.

Given Data / Assumptions:

• Quadratic: 7y^2 − 50y + k = 0
• Root: y = 7

Concept / Approach:
Direct substitution: set 7*(7^2) − 50*7 + k = 0 and solve for k. Keep arithmetic neat to avoid sign errors.

Step-by-Step Solution:
Compute: 7*(49) − 50*7 + k = 0343 − 350 + k = 0 ⇒ −7 + k = 0Therefore, k = 7

Verification / Alternative check:
Substitute back: 7*(49) − 350 + 7 = 343 − 343 = 0, confirming y = 7 is indeed a root.

Why Other Options Are Wrong:
1, 50/7, 7/50, 0 do not satisfy the zero condition upon substitution.

Common Pitfalls:
Arithmetic slips when multiplying 50*7 or 7*49. Double-check the sign before solving.

Final Answer:
7