RandomMath

Solving Quadratic Equations Solving Quadratic Equations "We do not learn mathematics, we just get used to it" - Amankwah Rawley A Quadratic Equation is a polynomial equation of the form \(ax^2 + bx + c = 0\) (A polynomial of order 2). There are several ways to solve quadratic equations. This article will elaborate on three methods. They are: Factorization Completing of Squares Quadratic Formula 1. Factorization Given a quadratic equation \(ax^2 + bx + c = 0\). The factorization method says Find the factors of c (including negative values). Find two factors of c such that they multiply to give c and sum to give b. Substitute the two factors for b Group the terms in Twos Then Factorize 1.1. Example Consider the following quadratic equation: \(x^2 + 5x + 6 = 0\) Firstly we recognize that \(a=1, b=5, c=6\) Using the Factorization method Find the factors of the c=6 (Factors simply means integers which can divide c exactly with...

 Proof of Quotient Rule: Let  \[y(x) = \dfrac{u(x)}{v(x)} \] Taking the natural Logarithm \[\ln y(x) = \ln \dfrac{u(x)}{v(x)}\] \[\ln y(x) = \ln u(x) - \ln v(x) \] Taking the Differential  \[ \dfrac{y'(x)}{y(x)} = \dfrac{u'(x)}{u(x)} - \dfrac{v'(x)}{v(x)} \] Multiplying by y(x) \[ y'(x) = y(x) (\dfrac{u'(x)}{u(x)} - \dfrac{v'(x)}{v(x)}) \] But \( y(x) = \dfrac{u(x)}{v(x)} \) \[ y'(x) = \dfrac{u(x)}{v(x)}\dfrac{u'(x)}{u(x)} -  \dfrac{u(x)}{v(x)}\dfrac{v'(x)}{v(x)}) \] \[ y'(x) = \dfrac{u'(x)}{v(x)} - \dfrac{u(x)v'(x)}{v(x)^2} \] \[ y'(x) = \dfrac{u'(x)v(x) - u'(x)v'(x)}{v(x)^2} \]

 Product Rule Proof: let  \[y(x) = u(x) \cdot v(x) \] Taking the natural log on both sides \[ \ln y(x) = \ln (u(x)\cdot v(x)) \] \[ \ln y(x) = \ln u(x) + \ln v(x) \] Taking the Differential \[\dfrac{1}{y(x)} y'(x) = \dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x) \] \[ y'(x) = y(x) ( \dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x)) \] But \(y(x) = u(x) \cdot v(x)\) \[ y'(x) = u(x) \cdot v(x) (\dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x)) \] \[ y'(x) = v(x) \cdot u'(x) + u(x) \cdot v'(x)) \]