Maths test 2

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

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Indices

a^m x a^n = a^m+n
(a^m)^n =a^mn
a^1/n = ^n√a
a^m ÷ a^n = a^m-n
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

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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
(x-3)^2 - 2
(the min. value of x^2 -6x +7 is -2 (since the min. value of (x-3)^2 is 0).

Factorising:
look for possible factors and remember correct signs.

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Solving quadratic equations

by Factorising.

By completeing the square:
x^2 - 6x -4 = 0
=> x^2 - 6x =4
=> x^2 6x +9 = 4 + 9
=> (x-3)^2 = 13
x-3 = +/-√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

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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.

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Quadratic graphs

if a > 0 the parabola will be the right way up

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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

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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.

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SImultaneous equations

Two linear:
3x-2y =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

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Coordinates of a straight line

Distance between two points: p(a1, b1) and q(a2,b2)
√(a1-a2)^2+(b1-b2)^2

Gradient: gradient of pq is m=b2-b1/a2-a1

Equation of a line: y=mx+c
equation of line that goes throught point (x1,y1) with gradient m y-y1 = m(x-x1)
or equation of line that goes through points (x1,y1) (x2,y2) is
y-y1/x-x1 = y2-y1/x2-x1

Perpendicular and parallel lines
Parallel if they have the same gradient
and perpendicular if the product of the gradient is the negative reciprocal (-1).

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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=√(a2-a1)^2 + (b2-b1)^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 

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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 .

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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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