Derivation of the Quadratic Formula Step by Step
Formulas
For the quadratic equation $ax^{2}+bx+c=0$ (where $a\neq 0$): $$ x=\dfrac{ -b\pm \sqrt { b^{2}-4ac } }{2a} $$
Explanation
Given a quadratic equation, its roots can be easily found through the formula.
Derivation
Strategy: The key to deriving the formula is to convert it into a complete square form. This is explained in great detail for children who are not familiar with math. Simply follow along without questioning, and try to replicate it multiple times.
$$ \begin{align*} && ax^{2} + bx + c =& 0 \\ \implies && \ ax^{2} + bx =& -c \\ \implies && x^{2} + \dfrac{b}{a}x =& -\dfrac{c}{a} \\ \implies && x^{2} + \dfrac{b}{a}x + \left( \dfrac{ b^{2} }{ 4a^{2} }-\dfrac{ b^{2} }{ 4a^{2} } \right) =& -\dfrac{c}{a} \text{(완전제곱꼴을 만들기 위한 트릭)} \\ \implies && \left( x^{2}+\dfrac{b}{a}x+\dfrac{ b^{2} }{ 4a^{2} } \right) -\dfrac{ b^{2} }{ 4a^{2} } =& -\dfrac{c}{a} \\ \implies && \left( x^{2}+\dfrac{b}{a}x+\dfrac{ b^{2} }{ 4a^{2} } \right) =& \dfrac{ b^{2} }{ 4a^{2} }-\dfrac{c}{a} \\ \implies && \left( x^{2}+\dfrac{b}{a}x+\dfrac{ b^{2} }{ 4a^{2} } \right) =& \dfrac{ b^{2} }{ 4a^{2} }-\dfrac{ 4ac }{ 4a^{2} } \\ \implies && \left( x^{2}+\dfrac{b}{a}x+\dfrac{ b^{2} }{ 4a^{2} } \right) =& \dfrac{ b^{2}-4ac }{ 4a^{2} } \\ \implies && \left( x+\dfrac{b}{2a} \right) \left( x+\dfrac{b}{2a} \right) =& \dfrac{ b^{2}-4ac }{ 4a^{2} } \\ \implies && { \left( x+\dfrac{b}{2a} \right) }^{ 2 } =& \dfrac{ b^{2}-4ac }{ 4a^{2} } \\ \implies && \left( x+\dfrac{b}{2a} \right) =& \pm \sqrt { \dfrac{ b^{2}-4ac }{ 4a^{2} } } \text{(양변에 루트를 취함)} \\ \implies && x+\dfrac{b}{2a} =& \pm \sqrt { \dfrac{ b^{2}-4ac }{ 4a^{2} } } \\ \implies && x+\dfrac{b}{2a} =& \pm \dfrac{ \sqrt { b^{2}-4ac } }{2a} \\ \implies && x =& -\dfrac{b}{2a}\pm \dfrac{ \sqrt { b^{2}-4ac } }{2a} \\ \implies && x =& \dfrac{ -b\pm \sqrt { b^{2}-4ac } }{2a} \end{align*} $$
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