In \(\bigtriangleup\)PQR, PQ = 10cm, QR = 8cm and RP = 6cm, the perpendicular RS is drawn from R to PQ. Find the length of RS

marlianlawa

2 years ago

To find the values of \( x \) for which \( (x^2 - 1) > 0 \), we need to solve the inequality \( x^2 - 1 > 0 \).

First, let's factor the expression:

\[ x^2 - 1 = (x + 1)(x - 1) \]

Now, we can identify the critical points by setting each factor equal to zero:

1. \( x + 1 = 0 \) implies \( x = -1 \)
2. \( x - 1 = 0 \) implies \( x = 1 \)

These critical points divide the real number line into three intervals: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).

Now, we'll choose test points from each interval to determine the sign of \( (x^2 - 1) \):

1. Test point in \( (-\infty, -1) \): \( x = -2 \)
\( (-2 + 1)(-2 - 1) = (-1)(-3) = 3 > 0 \) so \( (x^2 - 1) > 0 \) in this interval.
2. Test point in \( (-1, 1) \): \( x = 0 \)
\( (0 + 1)(0 - 1) = (1)(-1) = -1 < 0 \) so \( (x^2 - 1) < 0 \) in this interval.
3. Test point in \( (1, \infty) \): \( x = 2 \)
\( (2 + 1)(2 - 1) = (3)(1) = 3 > 0 \) so \( (x^2 - 1) > 0 \) in this interval.

So, the solution to \( (x^2 - 1) > 0 \) is \( x \in (-\infty, -1) \cup (1, \infty) \).