The solution set of 16−x2≤∣x+4∣ is....
Explanation
The absolute value has the definition
∣x+4∣={x+4;for x≥−4−(x+4);for x<−4
Solve the inequality
For x≥−4
16−x2≤∣x+4∣
16−x2≤x+4
−x2−x+12≤0
(−x+3)(x+4)≤0
x=3∨x=−4
Create the number line
Number Line for x≥−4
Interval (−∞,−4]∪[3,∞) indicates regions with negative or zero sign.
−
+
−
−4
3
From the condition x≥−4 and the number line region above, the solution set is {x≤−4∨x≥3}.
For x<−4
16−x2≤∣x+4∣
16−x2≤−(x+4)
−x2+x+20≤0
(−x−4)(x−5)≤0
x=−4∨x=5
The number line
Number Line for x<−4
Interval (−∞,−4]∪[5,∞) indicates regions with negative or zero sign.
−
+
−
−4
5
From the condition x<−4 and the second number line region above, the solution set is {x<−4}.
Therefore the combined solution is
HP={x≤−4∨x≥3}∪{x<−4}
HP={x≤−4∨x≥3}