# Completing the Square: Two Geometrical Proofs

On this website I previously showed you why the formulas used to complete the square work – and how they can be used to derive formulas such as the quadratic equation. Now, I’ll be doing something different, but related… On this post, I’ll be showing you how to come up with the formulas (2 in total) used to complete the square – geometrically.

#### Completing the Square Formula (Derivation 1): In the diagram above, what we can see is that: ${ x }^{ 2 }+2\cdot \frac { b }{ 2 } x+{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }$

This means that: ${ x }^{ 2 }+bx+{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }\\ \\ \therefore \quad { x }^{ 2 }+bx={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$

#### Completing the Square Formula (Derivation 2): In the diagram above, what we can see is that: ${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }+2\cdot \frac { b }{ 2 } \left( x-\frac { b }{ 2 } \right) +{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }$

This means that: ${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }+b\left( x-\frac { b }{ 2 } \right) +{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { { b }^{ 2 } }{ 2 } +\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { 2{ b }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\\ \\ \therefore \quad { x }^{ 2 }-bx={ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$

I hope these geometrical proofs have helped you better understand why the formulas we use to complete the square are in existence. Thanks for reading! 😀