Algebra Cheat Sheet Basic Properties & Facts Arithmetic Operations
b = ab c c
ab + ac = a ( b + c )
a b = a b
a
a
c
a
b c
bc
c
a d + bc
a
d
bd
b
+ =
a−b c−d
=
ab + ac a
=
If a < b and c > 0 then ac < bc and
ac
If a < b and c < 0 then ac > bc and
b
c
ad ad − bc
d
bd
− =
b−a
a+b
d −c
c
= b + c,
Properties of Inequalities If a < b then a + c < b + c and a − c < b − c
a
b
c
c
= +
a b ad = c bc d
a≠0
n
a a
(a
n
a
= a n+m
m
m
) =a
a
nm
= an−m =
m
a
a0
= 1,
≥0
ab
=
a+b
−a = a
a b
≤
a
b
+b
c
>
b c b c
=
a a b
Triangle Inequality
Distance Formula If P1 = ( x1 , y1 ) and P2
1 a
c a
<
Properties of Absolute Value if a ≥ 0 a a = if a < 0 −a
Exponent Properties n
a
= ( x2 , y2 )
are two
points the distance between them is
m− n
a≠0
d ( P1 , P2 )
= ( x2 − x1 ) 2 + ( y2 − y1 ) 2
n
n
( ab ) = a a
−n
n
b
a = a n b bn
n
1
=
a
1
n
−n
a =b b a
a n
= bn an
−n n
am
n
a
m n
=a a
n
a
n
n
a
n
1 n
= nm a
n
n
ab a b
= n an b
=
= a, if n is odd odd = a , if n is even even
= an
= (a
Properties of Radicals
Complex Numbers
1 m
i=
)
n
= ( an )
1 m
−1
= −1 −a = i a , a ≥ 0 di) = a + c + ( b + d ) i ( a + bi ) + ( c + di ( a + bi ) − ( c + di) = a − c + ( b − d ) i b d + ( ad + bc) i ( a + bi ) ( c + di ) = ac − bd ( a + bi ) ( a − bi ) = a2 + b2 a + bi
=
i
2
a 2 + b2
Complex Modulus
n
a
bi ) = a − bi bi ( a + bi
n
b
bi ) ( a + bi b i ) = a + bi bi ( a + bi
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu http://tutorial.math.lamar.edu..
Complex Conjugate 2
© 2005 Paul Dawkins
Logarithms and Log Properties Definition y y = log b x is equiva ivalent to x = b
Logarithm Properties log b b = 1 log b 1 = 0
= x blog x = x r log b ( x ) = r log b x log b ( xy ) = log b x + log b y x log b = log b x − log b y y x
Example log 5 125 = 3 because 5 3
logb b
= 125
Special Logarithms ln x = log e x natural log
log x = log10 x
common log
where e = 2.718281828K
b
The domain of log b x is x > 0
Factoring and Solving Factoring Formulas 2 2 x − a = ( x + a ) ( x − a ) x
2
Quadratic Formula 2 Solve ax + bx + c = 0 , a ≠ 0
+ 2ax + a2 = ( x + a ) 2
x =
2
3
Square Root Property
If x 2
= ( x − a ) ( x2 + ax + a2 ) x 2 n − a 2 n = ( xn − an ) ( xn + an ) x 3 − a 3
x
− an = ( x − a ) ( xn−1 + axn −2 + L + an−1 )
p
+ an = ( x + a ) ( xn−1 − axn−2 + a2 xn−3 − L + an−1 )
p
n
Solve 2 x
2
=
p then x = ± p
Absolute Value Equations/Inequalities If b is a positive number p = b p = − b or p= b ⇒
If n is odd then, x n
− 4 ac
2
2
− 3ax 2 + 3a2 x − a3 = ( x − a )3 3 3 2 2 x + a = ( x + a ) ( x − ax + a ) x
b
2a If b − 4ac > 0 - Two real unequal solns. If b 2 − 4ac = 0 - Repeated real solution. If b 2 − 4ac < 0 - Two complex solutions.
− 2ax + a2 = ( x − a ) 2 2 x + ( a + b ) x + ab = ( x + a) ( x + b) 3 x 3 + 3ax 2 + 3a 2 x + a3 = ( x + a ) x
−b ±
b
⇒ ⇒
−b < p < b p < −b or
p>b
Completing the Square (4) Factor the left side
− 6 x − 10 = 0
2
x − 3 = 29 2 4
2
(1) Divide by the coefficient of the x 2 x − 3 x − 5 = 0 (2) Move the constant to the other side. 2 x − 3 x = 5 (3) Take half the coefficient of x, square it and add it to both sides 2
x
2
− 3x + − 3 = 5 + − 3 2 2
2
(5) Use Square Root Property x −
3
2 (6) Solve for x
= 5 + 9 = 29 4 4
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu http://tutorial.math.lamar.edu..
29
=±
x =
4 3 2
±
=±
29 2
29 2
© 2005 Paul Dawkins
Functions and Graphs Constant Function y = a or f ( x) = a
Parabola/Quadratic Function x = ay 2 + by + c g ( y ) = ay 2 + by + c
Graph is a horizontal line passing through the point ( 0, a ) .
The graph is a parabola that opens right if a > 0 or left if a < 0 and has a vertex
Line/Linear Function y = mx + b or f ( x ) = mx + b
at
Graph is a line with point ( 0, b ) and
Circle
slope m.
2
2
( x − h ) + ( y − k ) = r 2 Graph is a circle with radius r and center ( h, k ) .
Slope Slope of the line containing the two points ( x1 , y1 ) and ( x2 , y2 ) is m=
y2 − y1 x2 − x1
=
rise
Ellipse
run
( x − h )
y = mx + b Point – Slope form The equation of the line with slope m and passing through the point ( x1 , y1 ) is y = y1 + m ( x − x1 )
Parabola/Quadratic Function y = a ( x − h )
2
+k
f ( x ) = a ( x − h)
2
+ k
The graph is a parabola that opens up if a > 0 or down if a < 0 and has a vertex at ( h, k ) . Parabola/Quadratic Function 2 2 y = ax + bx + c f ( x ) = ax + bx + c
The graph is a parabola that opens up if a > 0 or down if a < 0 and has a vertex b b . − 2a , f − 2a
2
( y − k )
2
+ =1 a2 b2 Graph is an ellipse with center ( h, k )
Slope – intercept form The equation of the line with slope m and y-intercept ( 0, b ) is
at
b , − b . g − 2a 2a
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
− =1 2 2 a b 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 ± . a Hyperbola
( y − k )
2
( x − h)
2
− =1 2 2 b a 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 ± . a
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu http://tutorial.math.lamar.edu..
© 2005 Paul Dawkins
Common Algebraic Errors Error
2 0
≠0
and
2 0
Reason/Correct/Justification/Example
≠2
Division by zero is undefined!
−32 = −9 , ( −3) 2 = 9
−32 ≠ 9
( x2 )
3
a b+c 1
1
b
c
2 1+1 1 1 A more complex version of the previous error.
≠ x −2 + x −3
1
=
a + bx
≠ 1 + bx
a
= +
= 1+
bx
Make sure you distribute the “-“!
2
2
( x + a ) = ( x + a ) ( x + a) =
+ a2 ≠ x + a x + a ≠ x + a
5 = 25
2
n
( x + a ) ≠ x n + an and
n
x + a
≠ n x+n a
≠ ( 2 x + 2)2
2
( 2 x + 2 ) ≠ 2 ( x + 1)
2
=
32 + 4 2
≠
32
x2 + 2 ax + a2
+
42
= 3+ 4 = 7
See previous error. More general versions of previous three errors. 2 ( x + 1)
2
bx bx
−a ( x − 1) = − ax + a
( x + a ) ≠ x2 + a2
2 ( x + 1)
1 1
≠ + =2
a a a a Beware of incorrect canceling!
−a ( x − 1) ≠ − ax − a
x
= x 2 x 2 x 2 = x6
a
3
a
3
a
≠ +
+x a + bx
x
2
( x2 )
≠ x5
Watch parenthesis!
2
= 2 ( x2 + 2 x + 1) = 2 x2 + 4 x + 2
2
( 2 x + 2 ) = 4 x2 + 8 x + 4 Square first then distribute! See the previous example. You can not factor out a constant if there is a power on the parethesis! 1
− x 2 + a 2 ≠ − a
b c
≠
ab c
a b ac ≠ c
b
x
2
+ a2
− x + a = ( − x + a ) 2 2
2
2
2
Now see the previous error. a 1 a c ac a = = = 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 http://tutorial.math.lamar.edu..
© 2005 Paul Dawkins
