212k views
3 votes
Yuri thinks that x = a is a root of the function q(x) = 6x³ - 19x² - 15x - 28. Explain to Yuri why x = a cannot be a root.

a) q(a) = 0
b) q(a) < 0
c) q(a) > 0
d) q(a) ≠ 0

1 Answer

2 votes

Final answer:

Yuri must evaluate q(a) to determine if x = a is a root. If q(a) is non-zero, then x = a is not a root of the function q(x). The specific value of a is required for further calculations.

Step-by-step explanation:

To explain to Yuri why x = a cannot be a root of the function q(x) = 6x³ - 19x² - 15x - 28, we must evaluate q(a). If q(a) = 0, then x = a would indeed be a root. However, if q(a) ≠ 0, then x = a is not a root of the function. The options provided by Yuri suggest evaluating whether q(a) is positive, negative, or non-zero to determine if x = a can be a root.

The correct answer to Yuri's question depends on the value of q(a). We need to substitute the value of a into the function q(x) and calculate the result. If after the substitution q(a) results in any number other than zero, then option d) q(a) ≠ 0 is the correct answer because it shows that x = a is not a root.

Without the specific value of a, we cannot determine whether q(a) is greater than, less than, or equal to zero; therefore, we cannot select options a), b), or c) as the direct answer. Only plagiarism free content is provided here, and no calculation step can be completed without the value of a.

User Miwin
by
8.3k points