Algebra

Try values: For : LHS1: , LHS2: , OK. Also Similarly, ⇒ works too.

Given that and , we substitute these into the expression: Simplify the terms to find that the result is equal to .

The solution of the inequality system can be found by solving both inequalities and analyzing the solution intervals.
The solution for gives , and solving gives the final solution set as .

The expression simplifies using the identity for expanding squares and factoring the terms.
Upon simplification, we get the correct expression as .

From Vieta’s formulas, we know for the quadratic equation : We are asked to find , which can be written as: Using Vieta’s formulas: Thus, the correct expression is , which is the answer.

We solve the equation algebraically.
After solving the equation, we find that there are 6 real roots that satisfy the given equation.

The determinant of the matrix is calculated, and we are given that and satisfy the equation . This equation implies that both and are roots of the equation , meaning that they are cube roots of unity. These roots can be expressed as , , and , where is a primitive cube root of unity.

Given this, the determinant simplifies to , which indicates that the determinant is a function of the eigenvalues of the matrix, and it is raised to the third power. This suggests that the matrix has a characteristic structure tied to the properties of cube roots of unity.

Therefore, the simplified determinant of the matrix is .

This expression simplifies to . The solution is deduced from the equation based on its symmetry and the properties of square roots.

We are given the equation: Simplify the left-hand side: Since , the equation becomes: Now, solve for : Thus, the value of is .