Maths test 2
 Created by: Jemima Marsden
 Created on: 011116 04:33
Surds
To simplify:
√50 = √25x2 = √25 x √2 = 5√2
^3√40 = √3√8x5 = 2 x ^3√5
Manipulating surds:
√a x √b = √ab e.g √2 x √18 = √2x18 = √36 = 6
√a/√b = √a/b e.g √64/√16 = 8/4 = 2
Simplifying by expressing as aproduct of prime factors:
√968 = √(2^3x11^2)
= √(2^2x11^2) x √2
= 2 x 11√2
= 22√2
Difference of squares:
(√a  √b)(√a+√b) = a  b
Rationalising the denominator:
Rationalise = getting rid of surds
2+3√5/3√5 = 2+3√5/3√5 x 3 + √5/3 + √5
= 6+ 3√5√5 + 9√5 + 2V5/ 3^2  (√5)^2
= 21 + 11√5 / 4
Indices
a^m x a^n = a^m+n
(a^m)^n =a^mn
a^1/n = ^n√a
a^m ÷ a^n = a^mn
a^0 = 1
a^m/n = ^n√a^m = (^n√a)^m
a^n = 1/a^n
Example:
find x if 9^2x = 27^x+1
9= 3^2 27=3^3
9^2x = 27^x+1 => (3^2)^2x = (3^3)^x+1
3^4x = 3^3x+3
4x=3x +3 => x=3
Quatratic functions
Completing the square:
coefficiant of x^2 must be +1
halve the coefficiant of x, squaare it then add it and subtract it
example: x^2 6x +6
x^2  6x +(3)^2 9 + 7
(x3)^2  2
(the min. value of x^2 6x +7 is 2 (since the min. value of (x3)^2 is 0).
Factorising:
look for possible factors and remember correct signs.
Solving quadratic equations
by Factorising.
By completeing the square:
x^2  6x 4 = 0
=> x^2  6x =4
=> x^2 6x +9 = 4 + 9
=> (x3)^2 = 13
x3 = +/√13 => x = 3+/√13 = 0.61 0r 6.61
By using the formula:
x=b+/√b^2 4ac/2a
The discriminant:
b^2 4ac
it will have: two distinct roots => b^2  4ac >0, only one real root => b^2  4ac=0, no real roots => b^2  4ac<0
Miscellaneous quadratic equations
Example: Solve 3^2x – 10 × 3^x + 9 = 0
Solution:
Notice that 3^2x = (3^x ) ^2
so put y = 3^x to give y^2 – 10y + 9 = 0
⇒ (y – 9)(y – 1) = 0
⇒ y = 9 or y = 1
⇒ 3x = 9 or 3x = 1 ⇒ x = 2 or x = 0.
Quadratic graphs
if a > 0 the parabola will be the right way up


b^2  4ac > 0 b^2  4ac = 0 b^2  4ac < 0
2 distinct roots only 1 real root no real roots
if a <0 the parabola will be the wrong way up


b^2  4ac > 0 b^2  4ac = 0 b^2  4ac < 0
2 distinct roots only 1 real root no real roots
show value of the y axis and the x axis if it meets it.
if you have completed the square give the coordinates of the vertex.
SImultaneous equations
Two linear:
3x2y =4 1
4x + 7y = 15 2
(make the coefficiants of x or y equal then add or subtract to eliminate x or y (same signs subtract))
4 x 1 => 12x  8 y = 16
3 x 2 => 12x +21y = 45
29y = 29
=> y=1
substitute in y
and x=2
One linear and one quadratic:
substitue in equation one to equation two and solve
Coordinates of a straight line
Distance between two points: p(a1, b1) and q(a2,b2)
√(a1a2)^2+(b1b2)^2
Gradient: gradient of pq is m=b2b1/a2a1
Equation of a line: y=mx+c
equation of line that goes throught point (x1,y1) with gradient m yy1 = m(xx1)
or equation of line that goes through points (x1,y1) (x2,y2) is
yy1/xx1 = y2y1/x2x1
Perpendicular and parallel lines
Parallel if they have the same gradient
and perpendicular if the product of the gradient is the negative reciprocal (1).
Coordinate geometry 2
Midpoint: mid point of the line joining p(a1,b1) and q(a2,b2) is (1/2(a1+b1), 1/2(a2, b2))
distance between two points:
using pythagoras' theorum PQ=√(a2a1)^2 + (b2b1)^2
perpendicular lines
two lines with gradient m1 and m2 are perpendicular <=> m1 x m2 = 1
Example: Find the equation of the line through (1, –5) which is perpendicular to the lline with equation y = 2x – 3
Solution: The gradient of y = 2x – 3 is 2
=> Gradient of perpendicular line is – 1/2
=> equation of perpendicular line is y – – 5 = –1/2 (x – 1)
=> x + 2y + 9 = 0
Circle coordinate geometry
Centre at the origin:
Take any point, P, on a circle centre the origin and radius 5.
Suppose that P has coordinates (x, y)
Using Pythagoras’ Theorem we have x^ 2 + y ^2 = 5^2 => x ^2 + y ^2 = 25 which is the equation of the circle. and in general the equation of a circle centre (0, 0) and radius r is x^ 2 + y^ 2 = r ^2 .
Logarithms
y = 2^x is an exponential function and its inverse is the logarithm function y = log2x.
Remember that the graph of an inverse function is the reflection of the original graph in y = x.
Rules of logarithms:
log a x = y <=> x = a^ y log a xy = log a x + log a y
log a (x/y) = log a x – log a y log a x^n = n log a x
log a 1 = 0 log a a = 1
Changing the base of a logarithm:
logab = logcb/logca
A particular case:
logab = logbb/logba = 1/logba
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