Question

2. If y = V9 – 3x, determine the value of x for which y will be a real number.

Answer 1

Answer:

x ≤ 3

Step-by-step explanation:

The expression inside the radical ≥ 0, thus

solve

9 - 3x ≥ 0 ( subtract 9 from both sides )

- 3x ≥ - 9

Divide both sides by - 3, reversing the symol as a result of dividing by a negative quantity

x ≤ 3

Any value of x less than or equal to 3 ensures $\sqrt{9-3x}$ is real