A quadratic function is represented by the following equation:

${\displaystyle y=ax}$2${\displaystyle +bx+c}$
• ${\displaystyle ax}$2 = Quadratic Term
• ${\displaystyle bx}$ = Linear Term
• ${\displaystyle c}$ = Constant Term

## Solving Quadratic Functions by Factoring

1. ${\displaystyle x}$2 ${\displaystyle +7x+6}$

• Factor them: We get ${\displaystyle (x+6)(x+1)}$.
• The linear terms must add to make 7.
• The constant terms needs to multiply to make 6.
• Set them out as problems to solve:
• ${\displaystyle x+6=0}$ → x = -6
• ${\displaystyle x+1=0}$ → x = -1
• Your answers are ${\displaystyle -6}$ and ${\displaystyle -1}$.

2. ${\displaystyle x}$2 ${\displaystyle -9}$.

• Factor them: We get ${\displaystyle (x-3)(x+3)}$.
• Set them out as problems to solve:
• ${\displaystyle x+3=0}$ → x = -3
• ${\displaystyle x-3=0}$ → x = 3
• Your answers are ${\displaystyle 3}$ and ${\displaystyle -3}$.

3. ${\displaystyle 2x}$2${\displaystyle +12x+10}$

• Divide all of the terms by the GCF: 2: We get a new problem to deal with, which is ${\displaystyle x}$2${\displaystyle +6x+5}$.
• Factor them: We get ${\displaystyle 2(x+5)(x+1)}$.
• Set them out as problems to solve:
• ${\displaystyle x+5=0}$ → x = -5
• ${\displaystyle x+1=0}$ → x = -1
• Your answers are ${\displaystyle -5}$ and ${\displaystyle -1}$.

## Solving Quadratic Factors by Completing the Square

1. ${\displaystyle x}$2 ${\displaystyle +8x}$ __ = ${\displaystyle 19}$

• Take the Linear Term and divide it by ${\displaystyle 2}$: We get ${\displaystyle 4}$.
• We take this number, ${\displaystyle 4}$, and square it: We get ${\displaystyle 16}$.
• We add ${\displaystyle 16}$ to ${\displaystyle 19}$: We get ${\displaystyle 35}$.
• We now have: ${\displaystyle (x+4)}$2 ${\displaystyle =35}$.
• We square both sides: We get ${\displaystyle x+4=}$${\displaystyle 35}$.
• We minus 4 to the other side. Here is our answer.: ${\displaystyle x=-4}$± √${\displaystyle 35}$.

2. ${\displaystyle x}$2 ${\displaystyle +x}$ + ___

• Take the Linear Term and divide it by ${\displaystyle 2}$: We get ${\displaystyle {\tfrac {1}{2}}}$.
• We take this number, ${\displaystyle {\tfrac {1}{2}}}$, and square it: We get ${\displaystyle {\tfrac {1}{4}}}$.
• We have our answer: ${\displaystyle x}$2 ${\displaystyle +x}$ + ${\displaystyle {\tfrac {1}{4}}}$.

3. ${\displaystyle x}$2 ${\displaystyle +45}$ = ${\displaystyle 10x}$

• Rearrange this problem so that it matches the standard format for a quadratic equation: We switch the ${\displaystyle 45}$ and the ${\displaystyle 10x}$ around, forming our new problem: ${\displaystyle x}$2 ${\displaystyle -10x}$ = ${\displaystyle -45}$.
• Divide the Linear Tearm, ${\displaystyle bx}$, by ${\displaystyle 2}$: This gives us ${\displaystyle 5}$.
• Square the ${\displaystyle 5}$: This gives us ${\displaystyle 25}$.
• Add the ${\displaystyle 25}$ to ${\displaystyle -45}$: This brings our problem to (${\displaystyle x}$ ${\displaystyle -5)}$2 = ${\displaystyle -20}$.
• Square both sides of the problem: This brings us to ${\displaystyle x-5=}$i√${\displaystyle 20}$.
• Find the square root of ${\displaystyle 20}$ (don't forget the ${\displaystyle i}$) and then add ${\displaystyle 5}$ to the opposite side to find your answer: Our final answer is ${\displaystyle x=-5}$± ${\displaystyle 2i}$${\displaystyle 5}$.