Algebra Cheat Sheet

Logarithms and Log Properties Definition y = log b x is equivalent to x = b y Algebra Cheat Sheet Basic Properties & Fa

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Logarithms and Log Properties Definition y = log b x is equivalent to x = b y

Algebra Cheat Sheet Basic Properties & Facts Properties of Inequalities If a < b then a + c < b + c and a - c < b - c

Arithmetic Operations æ b ö ab aç ÷ = ècø c

ab + ac = a ( b + c ) æaö ç ÷ a èbø = c bc

a b < c c a b If a < b and c < 0 then ac > bc and > c c If a < b and c > 0 then ac < bc and

a ac = æbö b çc÷ è ø

a c ad + bc + = b d bd

a c ad - bc - = b d bd

a-b b-a = c-d d -c

a+b a b = + c c c æaö ç b ÷ ad è ø= æ c ö bc çd ÷ è ø

ab + ac = b + c, a ¹ 0 a

Properties of Absolute Value if a ³ 0 ìa a =í if a < 0 î-a a ³0 -a = a

a +b £ a + b

Exponent Properties an 1 = a n -m = m -n am a

a na m = a n+ m

(a )

n m

( ab ) a

-n

n

= a nm

a 0 = 1, a ¹ 0 n

=a b

1 = n a

æaö ç ÷ èbø

-n

n

bn æbö =ç ÷ = n a è aø

n m

n

a =a

m n

1 n

a = nm a

( ) = (a )

a = a

Properties of Radicals

2

Complex Numbers i = -1

1 m

n

n

1 m

i = -1

-a = i a, a ³ 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 + b2

n

ab = a b

a + bi = a + b

n

a na = b nb

( a + bi ) = a - bi Complex Conjugate 2 ( a + bi )( a + bi ) = a + bi

n

n

a n = a , if n is odd

n

a n = a , if n is even

n

2

For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.

2

) = r log

b

x

æxö log b ç ÷ = log b x - logb y è yø The domain of log b x is x > 0 Quadratic Formula Solve ax 2 + bx + c = 0 , a ¹ 0

x 2 + 2ax + a 2 = ( x + a )

2

x 2 - 2 ax + a 2 = ( x - a )

2

-b ± b 2 - 4 ac 2a If b 2 - 4ac > 0 - Two real unequal solns. If b 2 - 4ac = 0 - Repeated real solution. If b 2 - 4ac < 0 - Two complex solutions. x=

x 2 + ( a + b ) x + ab = ( x + a )( x + b )

x3 - a 3 = ( x - a ) ( x2 + ax + a 2 )

2

b logb x = x

Factoring Formulas x 2 - a 2 = ( x + a )( x - a )

d ( P1 , P2 ) =

+ ( y2 - y1 )

r

log b ( xy ) = log b x + logb y

log x = log 10 x common log where e = 2.718281828K

x3 + 3ax 2 + 3a 2 x + a 3 = ( x + a )

2

log b ( x

Special Logarithms ln x = log e x natural log

Distance Formula If P1 = ( x1, y1 ) and P2 = ( x2 , y2 ) are two points the distance between them is

n

a æaö ç ÷ = n b èbø 1 n =a a-n

n n

Triangle Inequality

( x2 - x1 )

log b b x = x

Example log 5 125 = 3 because 53 = 125

Factoring and Solving

a a = b b

ab = a b

Logarithm Properties log b b = 1 log b 1 = 0

Complex Modulus

x3 - 3ax2 + 3a 2 x - a 3 = ( x - a )

3

3

Square Root Property If x2 = p then x = ± p

x3 + a 3 = ( x + a ) ( x2 - ax + a 2 )

Absolute Value Equations/Inequalities If b is a positive number p =b Þ p = - b or p = b

x 2 n - a 2 n = ( x n - a 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 + an = ( x + a)( x

n -1

- ax

n -2

2 n -3

+a x

-L + a

n -1

Þ

-b < p < b

p >b

Þ

p < - b or

p>b

)

Completing the Square (4) Factor the left side

Solve 2 x 2 - 6 x - 10 = 0

2

2

(1) Divide by the coefficient of the x 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 æ 3ö æ 3ö x 2 - 3x + ç - ÷ = 5 + ç - ÷ = 5 + = 4 4 è 2ø è 2ø

© 2005 Paul Dawkins

p 0 or left if a < 0 and has a vertex æ æ b ö b ö at ç g ç - ÷ , - ÷ . è è 2a ø 2a ø

Line/Linear Function y = mx + b or f ( x ) = mx + b Graph is a line with point ( 0,b ) and slope m.

Circle 2 2 ( x - h) + ( y - k ) = r2 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 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

( x - h)

2

f ( x) = a ( x - h) + k 2

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 y = ax 2 + bx + c f ( x ) = ax 2 + bx + c The graph is a parabola that opens up if a > 0 or down if a < 0 and has a vertex æ b æ b öö at ç - , f ç - ÷ ÷ . è 2a è 2a ø ø

( 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.

( x - h)

-

(y -k)

( y - k)

=1

2

( x - h) = 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 ± . a

For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.

Division by zero is undefined!

-32 ¹ 9

-32 = -9 ,

(x

(x

)

2 3

¹ x5

a a a ¹ + b+c b c 1 ¹ x -2 + x-3 x2 + x3

( x + a)

2

x2 + a2 ¹ x + a

( x + a)

n

( -3)

2

= 9 Watch parenthesis!

= x2 x2 x 2 = x6

( x + a)

¹ x2 + a 2

x+a ¹

)

2 3

1 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 1 + bx a

x+ a

¹ x n + a n and

2 ( x + 1) ¹ ( 2 x + 2 )

2

( 2x + 2)

2

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 2

Reason/Correct/Justification/Example

2 2 ¹ 0 and ¹ 2 0 0

n

x+a ¹

n

x+n a

= ( x + a )( x + a ) = x 2 + 2 ax + a 2

2

5 = 25 = 3 2 + 4 2 ¹ 3 2 + 4 2 = 3 + 4 = 7 See previous error. More general versions of previous three errors. 2 ( x + 1) = 2 ( x2 + 2 x + 1) = 2 x 2 + 4 x + 2 2

Hyperbola 2

Error

- a ( x - 1) ¹ - ax - a

Ellipse

2

Parabola/Quadratic Function y = a ( x - h) + k

Common Algebraic Errors

Parabola/Quadratic Function x = ay2 + by + c g ( y ) = ay2 + by + c

2

© 2005 Paul Dawkins

2

¹ 2 ( x + 1)

( 2x + 2)

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! 1

- x2 + a2 ¹ - x2 + a2 a ab ¹ æbö c çc÷ è ø æaö ç ÷ ac èbø ¹ c b

- x2 + a2 = ( - x2 + a2 )2 Now see the previous error. æaö ç1÷ a æ a ö æ c ö ac = è ø = ç ÷ç ÷ = æ b ö æ b ö è 1 øè b ø b ç ÷ ç ÷ ècø ècø æaö æaö ç ÷ ç ÷ è b ø = è b ø = æ a öæ 1 ö = a ç ÷ç ÷ c æ c ö è b ø è c ø bc ç ÷ 1 è ø

For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.

© 2005 Paul Dawkins