7. If \( x=-1-i \sqrt{3} \) is one root of the equation \( x^{2}+2 x+p=0 \) where \( p \) is a real number, find the value of \( p \) and the other root of the equation.

To find the value of \( p \) and the other root, we start with the given root \( x = -1 - i\sqrt{3} \). Since the coefficients of the polynomial are real, the complex roots come in conjugate pairs. Therefore, the other root is \( x = -1 + i\sqrt{3} \). Let's denote the roots as \( r_1 = -1 - i\sqrt{3} \) and \( r_2 = -1 + i\sqrt{3} \). According to Vieta's formulas, the sum of the roots \( r_1 + r_2 \) can be calculated as: \[ r_1 + r_2 = (-1 - i\sqrt{3}) + (-1 + i\sqrt{3}) = -2 \] So, the sum of the roots \( s = -2 \). The product of the roots \( r_1 \cdot r_2 \) is given by: \[ r_1 \cdot r_2 = (-1 - i\sqrt{3})(-1 + i\sqrt{3}) = (-1)^2 - (i\sqrt{3})^2 = 1 + 3 = 4 \] According to Vieta's formulas, for the quadratic equation \( x^2 + 2x + p = 0 \): 1. The sum of the roots \( -\frac{b}{a} = -2 \) implies \( b = 2 \). 2. The product of the roots \( \frac{c}{a} = p = 4 \). Thus, the value of \( p \) is \( 4 \). Now, summarizing, we have: - \( p = 4 \) - The other root is \( -1 + i\sqrt{3} \).