Derivation of the Quadratic Formula Step by Step
📂Abstract Algebra Derivation of the Quadratic Formula Step by Step For the quadratic equation a x 2 + b x + c = 0 ax^{2}+bx+c=0 a x 2 + b x + c = 0 a ≠ 0 a\neq 0 a = 0 x = − b ± b 2 − 4 a c 2 a
x=\dfrac{ -b\pm \sqrt { b^{2}-4ac } }{2a}
x = 2 a − b ± b 2 − 4 a c
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.
a x 2 + b x + c = 0 ⟹ a x 2 + b x = − c ⟹ x 2 + b a x = − c a ⟹ x 2 + b a x + ( b 2 4 a 2 − b 2 4 a 2 ) = − c a (완전제곱꼴을 만들기 위한 트릭) ⟹ ( x 2 + b a x + b 2 4 a 2 ) − b 2 4 a 2 = − c a ⟹ ( x 2 + b a x + b 2 4 a 2 ) = b 2 4 a 2 − c a ⟹ ( x 2 + b a x + b 2 4 a 2 ) = b 2 4 a 2 − 4 a c 4 a 2 ⟹ ( x 2 + b a x + b 2 4 a 2 ) = b 2 − 4 a c 4 a 2 ⟹ ( x + b 2 a ) ( x + b 2 a ) = b 2 − 4 a c 4 a 2 ⟹ ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 ⟹ ( x + b 2 a ) = ± b 2 − 4 a c 4 a 2 (양변에 루트를 취함) ⟹ x + b 2 a = ± b 2 − 4 a c 4 a 2 ⟹ x + b 2 a = ± b 2 − 4 a c 2 a ⟹ x = − b 2 a ± b 2 − 4 a c 2 a ⟹ x = − b ± b 2 − 4 a c 2 a
\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*}
⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ ⟹ a x 2 + b x + c = a x 2 + b x = x 2 + a b x = x 2 + a b x + ( 4 a 2 b 2 − 4 a 2 b 2 ) = ( x 2 + a b x + 4 a 2 b 2 ) − 4 a 2 b 2 = ( x 2 + a b x + 4 a 2 b 2 ) = ( x 2 + a b x + 4 a 2 b 2 ) = ( x 2 + a b x + 4 a 2 b 2 ) = ( x + 2 a b ) ( x + 2 a b ) = ( x + 2 a b ) 2 = ( x + 2 a b ) = x + 2 a b = x + 2 a b = x = x = 0 − c − a c − a c ( 완전제곱꼴을 만들기 위한 트릭 ) − a c 4 a 2 b 2 − a c 4 a 2 b 2 − 4 a 2 4 a c 4 a 2 b 2 − 4 a c 4 a 2 b 2 − 4 a c 4 a 2 b 2 − 4 a c ± 4 a 2 b 2 − 4 a c ( 양변에 루트를 취함 ) ± 4 a 2 b 2 − 4 a c ± 2 a b 2 − 4 a c − 2 a b ± 2 a b 2 − 4 a c 2 a − b ± b 2 − 4 a c
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