Solve the inequality below. Give the solution using interval notation. Give all values as integers or reduced fractions. (Do not convert fractions to decimals.) \( x^{2}+x<0 \)

To solve the inequality \( x^{2}+x<0 \), we first factor the left side: \( x(x+1)<0 \). This product is less than zero when one factor is positive and the other is negative. The critical points to test are where \( x(x+1) = 0 \), which gives us \( x = 0 \) and \( x = -1 \). Now, we test the intervals defined by these critical points: 1. For \( x < -1 \), let’s say \( x = -2 \): \( (-2)(-1) = 2 \) (positive) 2. For \( -1 < x < 0 \), let’s say \( x = -0.5 \): \( (-0.5)(0.5) = -0.25 \) (negative) 3. For \( x > 0 \), let’s say \( x = 1 \): \( (1)(2) = 2 \) (positive) Thus, the solution to the inequality is \( -1 < x < 0 \). In interval notation, this is written as \( (-1, 0) \).