# Completing the square

It turns out that there are 2 formulas which you can use to complete the square…

FIRST FORMULA:

The first formula looks like this: ${ x }^{ 2 }+bx\quad =\quad { \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$

We can prove that this is true by expanding the expression on the right… $\quad { \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } \right) -{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ ={ x }^{ 2 }+\frac { b }{ 2 } x+\frac { b }{ 2 } x+{ \left( \frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ ={ x }^{ 2 }+2\cdot \frac { b }{ 2 } x\\ \\ ={ x }^{ 2 }+bx$

SECOND FORMULA:

The second formula looks like this: ${ x }^{ 2 }-bx\quad =\quad { \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ { \left( \frac { b }{ 2 } \right) } }^{ 2 }$

We can prove that this is true by expanding the expression on the right… ${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ { \left( \frac { b }{ 2 } \right) } }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } \right) -{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ ={ x }^{ 2 }-\frac { b }{ 2 } x-\frac { b }{ 2 } x+{ \left( \frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ ={ x }^{ 2 }+2\left( -\frac { b }{ 2 } x \right) \\ \\ ={ x }^{ 2 }-bx$

DERIVING THE QUADRATIC FORMULA FROM SCRATCH:

Completing the square will allow you to come up with the quadratic formula from absolute scratch. See the example below: $a{ x }^{ 2 }+bx+c=0\\ \\ a{ x }^{ 2 }+bx=-c\\ \\ \frac { a{ x }^{ 2 } }{ a } +\frac { bx }{ a } =-\frac { c }{ a } \\ \\ { x }^{ 2 }+\frac { b }{ a } x=-\frac { c }{ a }$

At this point, what we can say is that: ${ x }^{ 2 }+px\quad =\quad { \left( x+\frac { p }{ 2 } \right) }^{ 2 }-{ \left( \frac { p }{ 2 } \right) }^{ 2 }$

And in this case: $p=\frac { b }{ a }$ $\therefore \quad \frac { p }{ 2 } =\frac { b }{ a } \cdot \frac { 1 }{ 2 } =\frac { b }{ 2a }$

So… Continuing… ${ x }^{ 2 }+\frac { b }{ a } x=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-{ \left( \frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } +{ \left( \frac { b }{ 2a } \right) }^{ 2 }\\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } +\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } \cdot \frac { 4a }{ 4a } +\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } }$

*What we did above is give each expression to the right the same denominator, so that the right hand side of the equation could be simplified. If you’d like to know more about giving expressions the same denominator, watch this video. It’s related to what we did to the right hand side of the equation above. ${ \left( x+\frac { b }{ 2a } \right) }^{ 2 }=-\frac { 4ac }{ 4{ a }^{ 2 } } +\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } } \\ \\ x+\frac { b }{ 2a } =\pm \sqrt { \frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } } } \\ \\ x+\frac { b }{ 2a } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=-\frac { b }{ 2a } \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ \therefore \quad x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

An extra page which could be useful, when trying to derive the quadratic formula form absolute scratch: