Surds Flashcards
Simplifying surds
√ab = √a x √b
find two numbers that x together to get the surd making sure one is a square number.
simplify √18
√9 x √2 = 3√2
simplify √200
√100 x √2 = 10√2
simplify √48
√4 x √12 = 2√12 = 2 x √4 √3
= 2x 2 x√3
= 4 √3
however if you spot the larger square root that goes into the surd it will make the process quicker e.g √48 - √16 x √3 = 4√3
simplify √12/√300
Simplify each half first.
= √4 x √3 / √100 x √3
= 2√3 / 10√3 (√3 is a common factor in numerator and denominator so they can cancel)
= 2/10 = 1/5
√24 x √150
Deal with them individually first.
= √4 x √6 x √25 x √6
= 2√6 x 5√6 = 10 x 6 = 60
√20 + √180
√4 x √5 + √36 x √5
= 2√5 + 6 √5 = 8 √5
√63 - √28
√9 x √7 - √4 x √7 = 3√7 - 2√7 = √7
√108 + √72
√36 x √3 + √36 x √2
= 6√3 + 6√2 (as the number under the square roots are different we cannot simplify any further)
however we could do -
6(√3 + √2) as there is two 6
Expand: √3(√2 + 5)
√3(√2 + 5) = √6 + 5 √3
Expand: 6(√3 + √6)
= 6√3 + 6 √6
Expand: √5(8-√7)
8√5 - √35
Expand: √6(√15 - 2√2)
= √90 - 2√12
= √9 x √10 - 2 x √4 x √3
= 3√10 - 4√3
√12(√50 + 3√10)
first we can simplify the surds
2√3(5√2 + 3 √10) = 10√6 + 6√30
Expand: (2 + √2)(3 - √5)
When expanding any double brackets we can use a grid.
|3 |-√5|
2| | |
√2| | |
Then we multiply the one on the side by the ones on the top.
|3 |-√5 |
2| 6 | -2 √5 |
√2| 3√2 | -√10 |
= 6 - 2√5 + 3√2 - √10
Expand: (2 - √5)(2 + √5)
(Can use the use grid method, just times everything in first bracket by the other)
= -1
Expand: (3 + √2)(2 +√3)
(Use the grid method)
= 6 + 2√2 + 3 √3 + √6
Expand: (√2 + 1)(√3 - √5)
(use the grid method)
=√6 -√10 + √3 - √5
Expand: (2√3 + 3√5)(2√2 - 5 √3)
|2√2 |-5√3
2√3| 4√6 |-10x3 = -30|
3√5 | 6√10|- 15√15 |
= 4√6 - 30 + 6√10 - 15√15
Rationalising the Denominator
multiply the numerator and the denominator by the same surd(the same surd that you are trying to rationalise) We want to get rid of the surd in the denominator and place turn the numerator into a surd.
Rationalise 2/√3
2/√3 x √3/√3 = 2√3/3
Rationalise 10/√5
10/√5 x √5/√5 =10√5/5 = 2√5
Rationalise
9/2√3
9/2√3 x √3/√3 = 9√3/6 = 3√3/2
Rationalise 1/1 + √2
With questions like this when times by the surd we switch the + or - symbol to the opposite so the surds cancel each other out.
1/1+√2 x 1-√2/1-√2
= 1-√2/1-√2+√2 -2
= 1 -√2/ -1 = -1 + √2
Rationalise 2/√2+2
2/√2+2 x √2-2/√2-2
= 2√2 -4/2-2√2 +2√2 -4
= 2√2 -4/-2 = -√2 + 2
Rationalise 3/4-√5
3/4-√5 x 4+√5//4+√5
= 12+3√5/16+4√5-4√5-5
=12+3√5/11
Rationalise 1+√2/3-√2
1+√2/3-√2 x 3+√2/3+√2
= 3+√2+3√2+2/9+3√2-3√2-2
= 5+4√2/7
Rationalise 4+2√3/3+3√2
4+2√3/3+3√2 x 3-3√2/3-3√2
= 12-12√2+6√3-6√6/9-9√2+9√2-18
=12-12√2+6√3-6√6/-9
=-4+4√2-2√3+2√6/3
Triangle ABC has base length 6√2-2√5, with area 5.
Find the exact perpendicular height of the triangle.
1/2 x(6√2-2√5) x h =5
(3√2 - √5) x h = 5
h = 5/3√2 - √5 x 3√2 + √5/3√2 + √5
= 15√2+5√5/18 + 3√10 - 3√10-5
=15√2+5√5/13
Rationalise the denominator of 1/√2 + √3 - √5
This probably will not be in an exam but its a good tester of your surds knowledge
solve x√2 + 5 = x - √2
= √2 + 5 = x - x√2
√2 + 5 = x(1 - √2)
=> x = √2 + 5/ 1 -√2 X 1+ √2/1+√2 = √2 + 2 + 5 + 5√2/ 1 - 2
= 7 + 6√2/ - 1 = -7 - 6√2
solve √2(x-3) = 4(x + √2)
√2x - 3√2 = 4x + 4√2
√2x - 4x = 7√2
(√2 - 4)x = 7√2
x = 7√2/√2 - 4 x √2 + 4/√2 + 4
=14 + 28√2/2 - 16 = 14+28√2/-14
= - 1 - 2√2
Find the exact co-ordinates of intersection of the two lines:
y = √2x + √3 - 2 and
y = √3 x + √3 - √6
first make one equation equal the other.
√2x + √3 - 2 = √3x + √3 - √6
(put all x’s on one side of equation)
√2x - √3x = 2 - √6
(√2 - √3)x = 2 - √6
x = 2 - √6/ √2-√3
(rationalise the denomiantor)
x= 2 - √6/ √2 - √3 x √2+√3/√2 +√3
= 2√2 + 2√3 - 2√3 - 3√2/ 2 - 3
=-√2/-1 = √2
(sub into one of the equations,equ1)
y = √2 x √2 + √3 - 2
= 2 + √3 - 2 =√3
(√2,√3)
Prove that 1/3√2 - 4 > 2 ( > is greater then)
(rationalise the denominator)
1/3√2-4 x 3√2 + 4/ 3√2+4
= 3√2+4/18 - 16
= 3√2 + 4/2 = 3/2√2 + 2 > 2
as 3/2√2 is >0
