91Ó°ÊÓ

A quadratic equation is a second-degree polynomial equation in one variable. It has the form:
\[ax^2 + bx + c = 0\]. In the given exercise, the quadratic equation derived is \[x^2 + 4x + 3 = 0\]. Here, \(a\) is 1, \(b\) is 4, and \(c\) is 3. Quadratic equations can be solved using several methods such as factoring, completing the square, or the quadratic formula. For this exercise, factoring was used. The solutions to a quadratic equation are where the graph of the parabola intersects the x-axis.

Factoring involves expressing the quadratic equation as a product of its linear factors. It is especially useful when the equation can be easily decomposed into simpler expressions. For the equation \(x^2 + 4x + 3 = 0\), we look for two numbers that multiply to 3 (the constant term \(c\)) and add up to 4 (the coefficient of \(x\)). These numbers are 1 and 3. So, we can write:
\[ (x + 1)(x + 3) = 0 \]. This means the solutions are found by setting each factor equal to zero:
\(x + 1 = 0\) (giving \(x = -1\)) and
\(x + 3 = 0\) (giving \(x = -3\)). These are the solutions for the quadratic equation.

Exponential functions have the general form \(f(x) = a \times b^{x}\) where \(a\) is a constant, \(b\) is the base, and \(x\) is the exponent. These functions grow by constant proportions and are widely used to model growth and decay processes. In the exercise, we start with an exponential equation \(2^{x^2 + 4x} = \frac{1}{8}\). To solve, we first express \(\frac{1}{8}\) as \(2^{-3}\) because \(8 = 2^3\) and reciprocals change the sign of the exponent. This rewrites the equation with the same base:
\(2^{x^2 + 4x} = 2^{-3}\).
Since the bases are the same, the exponents must be equal, leading to the quadratic equation we've already seen:
\(x^2 + 4x = -3\). This manipulation shows how understanding exponential functions can help solve more complex equations.