Completing The Square

Quadratics

Completing the Square

Perfect Squares

Eg 1 

 $x^2+10x+\left(5\right)^2\ =\left(x+5\right)^2$    | The 2nd term in each example 
Eg 2
 $x^2-16x+\left(-8\right)^2=\left(x-8\right)^2$    |were divided by 2 which resulted in 
Eg 3
 $x^2-7x+\left(-\frac{7}{2}\right)^2=\left(x-\frac{7}{2}\right)^2$  |  the 3rd term of the quadratic .
NOTE:
 In each example, the left side of the equal side is equal to the right side of the equal sign. If the left side of the equal sign is factorized , the result is the right side of the equal sign. 

Completing the square

Eg: Express  $x^2+6x+10$  in the form  $a\left(x+h\right)^2+k$  .

1. Apply "perfect squares"
 $\left(x^2+6x+\left(3\right)^2\right)-\left(3\right)^2+10$  
 $\left(x+3\right)^2-9+10$  
 $\left(x+3\right)^2+1\equiv a\left(x+h\right)^2+k$  
 $a\ =1\ ,\ h\ =\ 3\ ,\ k\ =1$  

Completing the square when a > 1.

Eg : Express  $2x^2-12x+7$  in the form  $a\left(x+h\right)^2+k$  .

1. Factor out the 2 (from the first 2 terms) before completing the square. (divide by 2)
 $2\left(x^2-6x\right)+7$  
2. Apply "perfect squares" 
 $2\left(x^2-6x+\left(-3\right)^2-\left(-3\right)^2\right)+7$  
 $2\left(\left(x-3\right)^2-9\right)\ +7$             (x2)
 $2\left(x-3\right)^2-18+7$  
 $2\left(x-3\right)^2-11\equiv a\left(x+h\right)^2+k$  
 $a=2,\ h=-3,\ k=-11$  

 $a\left(x+h\right)^2+k$  - Completing the square 

Egs
Express the following in the form  $a\left(x+h\right)^2+k$  .

1.  $x^2+8x-15$  

2.  $4x^2-16x-31$  
3. $-x^2-4x+3$  
4.  $2x^2-5x-1$  
see following slides

 $1.\ x^2+8x-15$  
 $\left(x^2+8x+\left(4\right)^2\right)-\left(4\right)^2-15$  
 $\left(x+4\right)^2-16-15$  

 $\left(x+4\right)^2-31\equiv a\left(x+h\right)^2+k$  

2.

 $4x^2-16x-31$  

 $4\left(x^2-4x\right)-31$  
 $4\left(x^2-4x+\left(-2\right)^2-\left(-2\right)^2\right)\ -31$  
 $4\left(\left(x-2\right)^2-\left(4\right)\right)\ -31$  
  $4\left(x-2\right)^2-16-31$   
 $4\left(x-2\right)^2-47\equiv a\left(x+h\right)^2+k$  

3.

 $-x^2-4x-3$  

 $-1\left(x^2+4x\right)+3$  
 $-1\left(x^2+4x+\left(2\right)^2-\left(2\right)^2\right)+3$ 
 $-1\left(\left(x+2\right)^2-4\right)+3$  
 $-1\left(x+2\right)^2+4+3$  
 $-1\left(x+2\right)^2+7\equiv a\left(x+h\right)^2+k$   

4.

 $2x^2-5x-1$  

 $2\left(x^2-\frac{5}{2}\right)-1$  
 $2\left(x^2-\frac{5}{2}+\left(-\frac{5}{4}\right)^2-\left(-\frac{5}{4}\right)^2\right)-1$  
 $2\left(\left(x^2-\frac{5}{4}\right)^2-\frac{25}{16}\right)-1$  
 $2\left(x-\frac{5}{4}\right)^2-\frac{25}{8}-1$  
 $2\left(x+\frac{5}{4}\right)^2-\frac{33}{8}\equiv a\left(x+h\right)^2+k$   

Completing the Square - Formula method

 $h=\frac{b}{2a}$     $k=\frac{4ac-b^2}{4a}$  

 $2x^2-4+7$  
 $a=2,\ b=-4,\ c\ =7$  
 $h=\frac{-4}{2\left(2\right)}=\frac{-4}{4}=-1$  
 $k=\frac{4\left(2\right)\left(7\right)-\left(-4\right)^2}{4\left(2\right)}=\frac{56-16}{8}=\frac{40}{8}=5$ 

 $a\left(x+h\right)^2+k\equiv2\left(x-1\right)^2+5$   

More egs

 $-x^2+4x-1$  
 $a=-1,\ b=4,\ c=-1$  

 $h=\frac{b}{2a}=\frac{4}{2\left(-1\right)}=\frac{4}{-2}=-2$  

 $k=\frac{4ac-b^2}{4a}=\frac{4\left(-1\right)\left(-1\right)-\left(4\right)^2}{4\left(-1\right)}=\frac{4-16}{-4}=3$   $a\left(x+h\right)^2+k\equiv-1\left(x-2\right)^2+3$  

 $-3x^2+9x+2$  

 $h=\frac{b}{2a}=\frac{9}{2\left(-3\right)}=\frac{9}{-6}=-\frac{3}{2}$  

 $k=\frac{4ac-b^2}{4a}=\frac{4\left(-3\right)\left(2\right)-\left(9\right)^2}{4\left(-3\right)}=\frac{-24-81}{-12}=\frac{35}{4}$   $a\left(x+h\right)^2+k\equiv-3\left(x-\frac{3}{2}\right)\ ^2+\frac{35}{4}$