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 \(y(x) = u(x) \cdot v(x)\) \[ y'(x) = u(x) \cdot v(x) (\dfrac{1}{u(x)} u'(x) + \dfrac{1}{v(x)} v'(x)) \] \[ y'(x) = v(x) \cdot u'(x) + u(x) \cdot v'(x)) \]

 Differential of \(\ln f(x)\) let \(y = \ln f(x)\) Taking exponential on both sides: \[e^{y} = e^{\ln f(x)}\] but \(e^{\ln f(x)} = f(x)\) since e and ln are inverse function of each other \[e^{y} = f(x)\] Differentiating both sides with respect to x (Remember y is a function of x thus when we differentiate y we ought to multiply with its derivative) \[e^{y} \dfrac{dy}{dx} = f'(x)\] \[\dfrac{dy}{dx} = \dfrac{f'(x)}{e^y}\] but \(e^y = f(x)\) \[\dfrac{dy}{dx} = \dfrac{f'(x)}{f(x)}\]

  Differential of \( e^{ax}\) let \( y = e^{ax}\) Taking the natural log on both sides: \[\ln y = \ln e^{ax}\] Apply one of the logarithm rules: \[\ln y = ax \ln e \] but \[\ln e = 1 \] \[\ln y = ax\] Taking the differential with respect to x: (Remember: y is a function of x therefore after differentiating we multiply by the derivative) \[\dfrac{1}{y} \dfrac{dy}{dx} = a \] \[\dfrac{dy}{dx} = ay\] but \(y = e^{ax}\) \[\dfrac{dy}{dx} = ae^{ax}\]  

 Formula Derivation of the Sum of the first n integers. First Method: Suppose we need to find the sum of the first n positive integers. Then, this can be written mathematically as: \[ S_{n} = \sum_{i=1}^{n} i \] \[ S_{n} = 1 + 2 + 3 + \cdot \cdot \cdot + n \]  but the above expression can also be written as (backwards): \[ S_{n} = n + (n-1) + (n-2) + \cdot \cdot \cdot + 1 \] Summing the two equations above:  \[2S_{n} = (n+1) + (n+1) + ... + (n+1) \] \[2S_{n} = n (n+1) \] \[ S_{n} = \dfrac{n(n+1)}{2} \]   Second Method: \[ S_{n} = 1 + 2 + 3 + \cdot \cdot \cdot + n \] Writing in sequence form: \[ S_{1} = 1  \] \[ S_{2} = 1 + 2 = 3 \] \[ S_{3} = 1 + 2 + 3 = 6 \] \[ S_{4} = 1 + 2 + 3 + 4 = 10 \] \[ \cdot \] \[ \cdot \] \[ \cdot \] \[ Seq = 1, 3, 6, 10, 15, ..., \sum_{i=1}^{n} i  \]  Note that the difference between each consecutive sequence is: \[ 1^{st}difference = 2, 3, 4, 5, 6, \cdot \cdot \cdot \] Since the difference is not constant, we take t...