Graphing - Absolute Value and Circle

Question

The diagram Shows The Graphs of $y = |x + 2|$ and $y = \sqrt{4 - x^2}$

Write down the solution for $\sqrt{4 - x^2}$ is equal to or less than $y = |x + 2|$.

Answer 1

Hints: you need to solve the inequality

$$\sqrt{4-x^2}\le |x+2|\implies 4-x^2\le x^2+4x+4\implies x^2+2x\ge 0$$

Remember that when you solve inequalities by squaring sides, you must check each and every "solution" you get at the end in the original inequality (why?)

Answer 2

HINT: You are asked to write down the values of x for which $\sqrt{4-x^2}\le|x+2|$. What would you do to remove square root and mod?