Completing The Square (Why It Works)

Prove that:

{ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }-bx

Proof:

{ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } \right) -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }-2x\frac { b }{ 2 } +\frac { { b }^{ 2 } }{ 4 } -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }-bx

Second Proof:

As:\\ \\ { p }^{ 2 }-{ q }^{ 2 }=\left( p+q \right) \left( p-q \right) \\ \\ And:\\ \\ p=\left( x-\frac { b }{ 2 } \right) \\ \\ And:\\ \\ q=\frac { b }{ 2 } ,\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } +\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } -\frac { b }{ 2 } \right) \\ \\ =x\left( x-2\frac { b }{ 2 } \right) \\ \\ =x\left( x-b \right) \\ \\ ={ x }^{ 2 }-bx

Prove that:

{ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }+bx

Proof:

{ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } \right) -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }+2x\frac { b }{ 2 } +\frac { { b }^{ 2 } }{ 4 } -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }+bx

Second Proof:

As:\\ \\ { p }^{ 2 }-{ q }^{ 2 }=\left( p+q \right) \left( p-q \right) \\ \\ And:\\ \\ p=\left( x+\frac { b }{ 2 } \right) \\ \\ And:\\ \\ q=\frac { b }{ 2 } ,\\ \\ { \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } +\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } -\frac { b }{ 2 } \right) \\ \\ =\left( x+2\frac { b }{ 2 } \right) x\\ \\ =x\left( x+b \right) \\ \\ ={ x }^{ 2 }+bx

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