Question 6

If $$y^2$$ + 3y - 18 ≥ 0, which of the following is true?

Solution

$$y^2 + 3y - 18 \geq 0$$

$$\Rightarrow y^2 + 6y - 3y - 18 \geq 0$$

$$\Rightarrow y (y+6) - 3 (y+6) \geq 0$$

$$\Rightarrow (y-3) (y+6) \geq 0$$

$$\Rightarrow y\geq3 and y\leq-6$$

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