Solve the inequality. Write the solution set in interval notation, if possible. Simplify your answer, if necessary.

[tex]\[ 16z^2 - 25 \ \textless \ 0 \][/tex]

The solution set is [tex]$\square$[/tex] [tex]$\square$[/tex].



Answer :

To solve the inequality [tex]\( 16z^2 - 25 < 0 \)[/tex], let's proceed step-by-step.

1. Factor the quadratic expression:

Start by rewriting the inequality:
[tex]\[ 16z^2 - 25 < 0 \][/tex]

Notice that [tex]\( 16z^2 - 25 \)[/tex] is a difference of squares, which can be factored as follows:
[tex]\[ 16z^2 - 25 = (4z - 5)(4z + 5) \][/tex]

So, the inequality becomes:
[tex]\[ (4z - 5)(4z + 5) < 0 \][/tex]

2. Find the critical points:

The critical points are the values of [tex]\( z \)[/tex] where each factor equals zero:
[tex]\[ 4z - 5 = 0 \quad \Rightarrow \quad z = \frac{5}{4} \][/tex]
[tex]\[ 4z + 5 = 0 \quad \Rightarrow \quad z = -\frac{5}{4} \][/tex]

These points divide the number line into three intervals:
[tex]\[ (-\infty, -\frac{5}{4}), \quad (-\frac{5}{4}, \frac{5}{4}), \quad (\frac{5}{4}, \infty) \][/tex]

3. Test the sign of each interval:

We'll test a point from each interval to determine where our inequality holds true.

- For [tex]\( z \)[/tex] in [tex]\( (-\infty, -\frac{5}{4}) \)[/tex]:
Choose [tex]\( z = -2 \)[/tex]:
[tex]\[ (4(-2) - 5)(4(-2) + 5) = (-8 - 5)(-8 + 5) = (-13)(-3) = 39 \quad (\text{positive}) \][/tex]

- For [tex]\( z \)[/tex] in [tex]\( (-\frac{5}{4}, \frac{5}{4}) \)[/tex]:
Choose [tex]\( z = 0 \)[/tex]:
[tex]\[ (4(0) - 5)(4(0) + 5) = (-5)(5) = -25 \quad (\text{negative}) \][/tex]

- For [tex]\( z \)[/tex] in [tex]\( (\frac{5}{4}, \infty) \)[/tex]:
Choose [tex]\( z = 2 \)[/tex]:
[tex]\[ (4(2) - 5)(4(2) + 5) = (8 - 5)(8 + 5) = (3)(13) = 39 \quad (\text{positive}) \][/tex]

The inequality [tex]\( (4z - 5)(4z + 5) < 0 \)[/tex] holds true in the interval where the product is negative, which is [tex]\( (-\frac{5}{4}, \frac{5}{4}) \)[/tex].

4. Write the solution in interval notation:

The solution set in interval notation is:
[tex]\[ (-\frac{5}{4}, \frac{5}{4}) \][/tex]

Thus, the solution set to the inequality [tex]\( 16z^2 - 25 < 0 \)[/tex] is [tex]\( \boxed{(-\frac{5}{4}, \frac{5}{4})} \)[/tex].