+0

# Algebra

0
1
1
+212

Solve the inequality $4t^2 \le -9t + 12 - 14t + 36.$ Write your answer in interval notation.

Jul 31, 2024

#1
+1485
+1

First, let's move all the terms to one side and combine all like terms.

We have the inequality

$$4t^2-48+23t\le \:0$$

Now, let's complete the square for t. We get

$$4\left(t+\frac{23}{8}\right)^2-\frac{1297}{16}\le \:0$$

This format allows us to isolate t in a much simpler manner. Now, let's start the process. We get

$$4\left(t+\frac{23}{8}\right)^2\le \:\frac{1297}{16}$$

$$\left(t+\frac{23}{8}\right)^2\le \:\frac{1297}{64}$$

Now, square rooting both sides, it gives us two parameters to focus on for t. We get

$$-\sqrt{\frac{1297}{64}}\le \:t+\frac{23}{8}\le \:\sqrt{\frac{1297}{64}}\\ \frac{-\sqrt{1297}-23}{8}\le \:t\le \:\frac{\sqrt{1297}-23}{8}$$

Now, the final step is to write this in interval notation. We finally get that t is $$[\frac{-\sqrt{1297}-23}{8},\frac{\sqrt{1297}-23}{8}]$$

Thanks! :)

Jul 31, 2024
edited by NotThatSmart  Jul 31, 2024