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diff --git a/Content/posts/2024-03-26-Derivation-of-the-Quadratic-Equation.md b/Content/posts/2024-03-26-Derivation-of-the-Quadratic-Equation.md new file mode 100644 index 0000000..0435a6c --- /dev/null +++ b/Content/posts/2024-03-26-Derivation-of-the-Quadratic-Equation.md @@ -0,0 +1,55 @@ +--- +date: 2024-03-26 15:36 +description: Quick derivation of the quadratic equation by completing the square +tags: mathematics +--- + +# Quadratic Formula Derivation + +The standard form of a quadratic equation is: + +$$ +ax^2 + bx + c = 0 +$$ + +Here, $a, b, c \in \mathbb{R}$, and $a \neq 0$ + +We begin by first dividing both sides by the coefficient $a$ + +$$ +\implies x^2 + \frac{b}{a}x + \frac{c}{a} = 0 +$$ + +We can rearrange the equation: + +$$ +x^2 + \frac{b}{a}x = - \frac{c}{a} +$$ + +We can then use the method of completing the square. ([Maths is Fun](https://www.mathsisfun.com/algebra/completing-square.html) has a really good explanation for this technique) + +$$ +x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 = \frac{-c}{a} + (\frac{b}{2a})^2 +$$ + +On our LHS, we can clearly recognize that it is the expanded form of $(x + d)^2$ i.e $x^2 + 2x\cdot d + d^2$ + +$$ +\implies (x + \frac{b}{2a})^2 = \frac{-c}{a} + \frac{b^2}{4a^2} = \frac{-4ac + b^2}{4a^2} +$$ + +Taking the square root of both sides + +$$ +\begin{align*} +x + \frac{b}{2a} &= \frac{\sqrt{-4ac + b^2}}{2a} \\ +x &= \frac{\pm \sqrt{-4ac + b^2} - b}{2a} \\ +&= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} +\end{align*} +$$ + +This gives you the world famous quadratic formula: + +$$ +x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} +$$ |