Problem 57 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve for \(x\) $$\left|\begin{array}{rr} x+3 & 2 \\ 1 & x+2 \end{array}\right|=0$$

Step-by-Step Solution

Answer

The solutions for \(x\) are -1 and -4.

Step 1: Calculate the Determinant

The determinant of a 2x2 matrix is calculated using the formula: \(\text{{Det(A)}} = a \cdot d - b \cdot c\), where a, b, c, and d are elements of the matrix. In this case, it will be calculated as: \((x+3) \cdot (x+2) - (1 \cdot 2)\)

Step 2: Set the Determinant Equals to Zero

Since it is given that the determinant equals zero, the equation then becomes: \((x+3) \cdot (x+2) - 2 = 0\)

Step 3: Simplify and Solve for \(x\)

By performing the multiplication and minus operation, the equation simplifies to: \(x^2 + 5x + 6 - 2 = 0\). This further simplifies to \(x^2 + 5x + 4 = 0\). By applying the quadratic formula \(-b \pm \sqrt {b^2 - 4ac} \over 2a\), you find the solutions: \(x = -4\) and \(x = -1\).

Key Concepts

Simplifying Algebraic Expressions

The process of simplifying algebraic expressions makes them easier to understand and work with. By combining like terms and using the distributive property, expressions can often be reduced to a more manageable form. In the case of our exercise, simplifying the multiplication within the determinant leads to a quadratic equation that can then be solved using the quadratic formula.