91Ó°ÊÓ

A quadratic equation is a polynomial equation of the form defined as \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constant coefficients. To solve quadratic equations, we commonly use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\). This formula works for any quadratic equation, whether it has real or complex solutions. In our problem, the quadratic equation given is: \(x^2 - 6x + 25 = 0\). Here, \(a = 1, b = -6,\) and \(c = 25\). To find the solutions, we need to calculate the discriminant \(\Delta = b^2 - 4ac\). Here, \(\Delta = (-6)^2 - 4\cdot 1\cdot 25 = 36 - 100 = -64\). Since the discriminant is a negative number, \(-64\), our solutions will be complex numbers. Plugging these into the quadratic formula gives us the roots: \(x = 3 \pm 4i\). These complex roots are \(3 + 4i\) and \(3 - 4i\).

Complex numbers are numbers that have both a real part and an imaginary part, usually written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The imaginary unit \(i\) satisfies the equation \(i^2 = -1\). Complex numbers are essential in finding solutions to certain types of quadratic equations, especially when the discriminant \(\Delta\) is less than zero, resulting in no real roots.In our exercise, the quadratic equation \(x^2 - 6x + 25\) gave us complex roots because \(\Delta = -64\). These roots are \(3 + 4i\) and \(3 - 4i\), meaning that even though the polynomial \(f(x) = x^2 - 6x + 25\) is not factorable using real numbers, it is factored in terms of complex numbers. This highlights both the generality and utility of complex numbers in algebra.

The Factor Theorem states that for a polynomial \(p(x)\), if \(p(c) = 0\), then \(x - c\) is a factor of \(p(x)\). This theorem helps us find the roots or zeros of the polynomial. In synthetic division, we use this theorem to test if a certain value makes the polynomial equal to zero, hence verifying if the corresponding binomial is a factor.In the exercise, we used synthetic division to check if \(x - (3 + 4i)\) and \(x - (3 - 4i)\) are factors of \(f(x) = x^2 - 6x + 25\). However, seeing non-zero remainders of \(50 - 6i\) and \(50 + 6i\), we determined that neither \(3 + 4i\) nor \(3 - 4i\) is a root of the polynomial, meaning that neither binomial is a factor. Despite this, solving the equation \(x^2 - 6x + 25 = 0\) with the quadratic formula confirmed that the actual roots are indeed \(3 + 4i\) and \(3 - 4i\).