# Algebra Cheat Sheet

```Algebra Cheat Sheet
Basic Properties &amp; Facts
Arithmetic Operations
Properties of Inequalities
If a &lt; b then a + c &lt; b + c and a − c &lt; b − c
a b
If a &lt; b and c &gt; 0 then ac &lt; bc and &lt;
c c
a b
If a &lt; b and c &lt; 0 then ac &gt; bc and &gt;
c c
 b  ab
a  =
c c
ab + ac = a ( b + c )
a
  a
b =
c
bc
a
ac
=
b b
 
c
+ =
b d
bd
− =
b d
bd
a −b b−a
=
c−d d −c
a+b a b
= +
c
c c
a
b =
 c  bc
 
d
ab + ac
= b + c, a ≠ 0
a
Properties of Absolute Value
if a ≥ 0
a
a =
if a &lt; 0
 −a
a ≥0
−a = a
a+b ≤ a + b
(a )
n m
an
1
= a n−m = m−n
m
a
a
( ab )
a 0 = 1, a ≠ 0
n
n
a −n =
a
 
b
= a nm
−n
1
an
n
bn
b
=  = n
a
a
n
m
1
a = an
m n
a = nm a
( x2 − x1 ) + ( y2 − y1 )
2
2
n
Complex Numbers
i = −1
( ) = (a )
a = a
n
d ( P1 , P2 ) =
a
a
  = n
b
b
1
= an
−n
a
= a nb n
Triangle Inequality
Distance Formula
If P1 = ( x1 , y1 ) and P2 = ( x2 , y2 ) are two
points the distance between them is
Exponent Properties
a n a m = a n+m
a
a
=
b
b
ab = a b
1
m
n
n
1
m
i 2 = −1
−a = i a , a ≥ 0
( a + bi ) + ( c + di ) = a + c + ( b + d ) i
( a + bi ) − ( c + di ) = a − c + ( b − d ) i
( a + bi )( c + di ) = ac − bd + ( ad + bc ) i
( a + bi )( a − bi ) = a 2 + b 2
n
ab = n a n b
a + bi = a 2 + b 2
n
a na
=
b nb
( a + bi ) = a − bi Complex Conjugate
2
( a + bi )( a + bi ) = a + bi
n
a n = a, if n is odd
n
a n = a , if n is even
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
Complex Modulus
&copy; 2005 Paul Dawkins
Logarithms and Log Properties
Definition
y = log b x is equivalent to x = b y
Logarithm Properties
log b b = 1
log b 1 = 0
log b b x = x
b logb x = x
log b ( x r ) = r log b x
Example
log 5 125 = 3 because 53 = 125
log b ( xy ) = log b x + log b y
Special Logarithms
ln x = log e x
natural log
x
log b   = log b x − log b y
 y
log x = log10 x common log
where e = 2.718281828K
The domain of log b x is x &gt; 0
Factoring and Solving
Factoring Formulas
x 2 − a 2 = ( x + a )( x − a )
Solve ax 2 + bx + c = 0 , a ≠ 0
x 2 + 2ax + a 2 = ( x + a )
2
x 2 − 2ax + a 2 = ( x − a )
2
−b &plusmn; b 2 − 4ac
2a
2
If b − 4ac &gt; 0 - Two real unequal solns.
If b 2 − 4ac = 0 - Repeated real solution.
If b 2 − 4ac &lt; 0 - Two complex solutions.
x=
x 2 + ( a + b ) x + ab = ( x + a )( x + b )
x3 + 3ax 2 + 3a 2 x + a 3 = ( x + a )
x3 − 3ax 2 + 3a 2 x − a 3 = ( x − a )
3
3
Square Root Property
If x 2 = p then x = &plusmn; p
x3 + a3 = ( x + a ) ( x 2 − ax + a 2 )
x3 − a 3 = ( x − a ) ( x 2 + ax + a 2 )
x −a
2n
2n
= (x −a
n
n
)( x
n
+a
n
)
If n is odd then,
x n − a n = ( x − a ) ( x n −1 + ax n − 2 + L + a n −1 )
xn + a n
Absolute Value Equations/Inequalities
If b is a positive number
p =b
⇒
p = −b or p = b
p &lt;b
⇒
−b &lt; p &lt; b
p &gt;b
⇒
p &lt; −b or
p&gt;b
= ( x + a ) ( x n −1 − ax n − 2 + a 2 x n −3 − L + a n −1 )
Completing the Square
(4) Factor the left side
Solve 2 x − 6 x − 10 = 0
2
2
(1) Divide by the coefficient of the x 2
x 2 − 3x − 5 = 0
(2) Move the constant to the other side.
x 2 − 3x = 5
(3) Take half the coefficient of x, square
it and add it to both sides
2
2
9 29
 3
 3
x 2 − 3x +  −  = 5 +  −  = 5 + =
4 4
 2
 2
3
29

x−  =
2
4

(5) Use Square Root Property
3
29
29
x− = &plusmn;
=&plusmn;
2
4
2
(6) Solve for x
3
29
x= &plusmn;
2
2
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
&copy; 2005 Paul Dawkins
Functions and Graphs
Constant Function
y = a or f ( x ) = a
Graph is a horizontal line passing
through the point ( 0, a ) .
Line/Linear Function
y = mx + b or f ( x ) = mx + b
Graph is a line with point ( 0, b ) and
slope m.
Slope
Slope of the line containing the two
points ( x1 , y1 ) and ( x2 , y2 ) is
y2 − y1 rise
=
x2 − x1 run
Slope – intercept form
The equation of the line with slope m
and y-intercept ( 0,b ) is
y = mx + b
Point – Slope form
The equation of the line with slope m
and passing through the point ( x1 , y1 ) is
m=
y = y1 + m ( x − x1 )
2
2
y = a ( x − h) + k
f ( x) = a ( x − h) + k
The graph is a parabola that opens up if
a &gt; 0 or down if a &lt; 0 and has a vertex
at ( h, k ) .
y = ax 2 + bx + c f ( x ) = ax 2 + bx + c
The graph is a parabola that opens up if
a &gt; 0 or down if a &lt; 0 and has a vertex
 b
 b 
at  − , f  −   .
 2a  2 a  
x = ay 2 + by + c g ( y ) = ay 2 + by + c
The graph is a parabola that opens right
if a &gt; 0 or left if a &lt; 0 and has a vertex
  b  b 
at  g  −  , −  .
  2a  2 a 
Circle
2
2
( x − h) + ( y − k ) = r 2
Graph is a circle with radius r and center
( h, k ) .
Ellipse
( x − h)
2
( y −k)
+
2
=1
a2
b2
Graph is an ellipse with center ( h, k )
with vertices a units right/left from the
center and vertices b units up/down from
the center.
Hyperbola
( x − h)
2
( y −k)
−
2
( x − h)
2
=1
a2
b2
Graph is a hyperbola that opens left and
right, has a center at ( h, k ) , vertices a
units left/right of center and asymptotes
b
that pass through center with slope &plusmn; .
a
Hyperbola
(y −k)
2
=1
b2
a2
Graph is a hyperbola that opens up and
down, has a center at ( h, k ) , vertices b
units up/down from the center and
asymptotes that pass through center with
b
slope &plusmn; .
a
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
−
&copy; 2005 Paul Dawkins
Common Algebraic Errors
Error
Reason/Correct/Justification/Example
2
2
≠ 0 and ≠ 2
0
0
Division by zero is undefined!
−32 ≠ 9
−32 = −9 ,
(x )
(x )
2 3
2 3
≠ x5
a
a a
≠ +
b+c b c
1
≠ x −2 + x −3
2
3
x +x
− a ( x − 1) ≠ − ax − a
x2 + a2 ≠ x + a
x+a ≠ x + a
( x + a)
≠ x n + a n and
n
= 9 Watch parenthesis!
= x2 x2 x2 = x6
( x + a)
≠ x2 + a2
2
2
1
1
1 1
=
≠ + =2
2 1+1 1 1
A more complex version of the previous
error.
a + bx a bx
bx
= +
= 1+
a
a a
a
Beware of incorrect canceling!
− a ( x − 1) = − ax + a
Make sure you distribute the “-“!
a + bx
≠ 1 + bx
a
( x + a)
( −3 )
n
x+a ≠ n x + n a
= ( x + a )( x + a ) = x 2 + 2ax + a 2
2
5 = 25 = 32 + 42 ≠ 32 + 42 = 3 + 4 = 7
See previous error.
More general versions of previous three
errors.
2 ( x + 1) = 2 ( x 2 + 2 x + 1) = 2 x 2 + 4 x + 2
2
2 ( x + 1) ≠ ( 2 x + 2 )
2
( 2 x + 2)
2
2
2
≠ 2 ( x + 1)
( 2 x + 2)
= 4 x2 + 8x + 4
Square first then distribute!
See the previous example. You can not
factor out a constant if there is a power on
the parethesis!
2
1
− x2 + a2 ≠ − x2 + a2
a
ab
≠
b c
 
c
a
  ac
b ≠
c
b
− x2 + a2 = ( − x2 + a 2 ) 2
Now see the previous error.
a
 
a
1
 a  c  ac
=   =    =
 b   b   1  b  b
   
c c
a a
   
 b  =  b  =  a  1  = a
  
c
 c   b  c  bc
 
1
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
&copy; 2005 Paul Dawkins
```