Answer:
[tex]1\leq x\leq \frac{21}{4}[/tex]
Step-by-step explanation:
First, we determine the domain of the inequality, knowing that it doesn't exist the result of a negative square root. So we have to consider that:
[tex]4x-4\geq 0\\4x\geq 4\\\\x\geq \frac{4}{4} \\\\x\geq 1 \\\\[/tex]
Then we isolate X to know the result of the inequality:
[tex]\sqrt{4x-4} +8 \leq 5\\4x-4+8\leq 5^{2} \\4x-4+8\leq 25\\4x+4\leq 25\\4x\leq 25-4\\\\4x\leq 21\\\\\\x\leq \frac{21}{4} \\\\[/tex]
The result is
[tex]1\leq x\leq \frac{21}{4}[/tex]
