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.

Given the quadratic equation: \[ x^2 + 2x + p = 0 \] and one of its roots is: \[ x = -1 - i\sqrt{3} \] Since the coefficients of the equation are real numbers, the complex roots must occur in conjugate pairs. Therefore, the other root must be the complex conjugate of the given root: \[ x = -1 + i\sqrt{3} \] To find the value of \( p \), we can use Vieta's formulas, which relate the coefficients of the polynomial to the sums and products of its roots. 1. **Sum of the roots:** \[ (-1 - i\sqrt{3}) + (-1 + i\sqrt{3}) = -2 \] This matches the coefficient of \( x \) in the equation \( x^2 + 2x + p = 0 \), which is indeed \( 2 \times (-1) = -2 \). 2. **Product of the roots:** \[ (-1 - i\sqrt{3})(-1 + i\sqrt{3}) = (-1)^2 - (i\sqrt{3})^2 = 1 - (-3) = 4 \] Thus, the product of the roots is equal to \( p \). **Final Answer:** - The value of \( p \) is **4**. - The other root of the equation is **\( -1 + i\sqrt{3} \)**. **Answer Summary:** After simplifying, p = 4 and the other root is –1 plus i times √3. Thus, p = 4 and the other root is −1 + i √3.