Download A6 Quadratic equations

SLO
Solving quadratic equations
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Words to learn
Factorise: put into brackets
Solve: find answer
Quadratic: x2 as highest power
Solution: answer (usually a number)
2 of 48
Easy trick
Think of any number, don’t tell me…..
Times by 10
Add 5
Times by 0
I bet you now have the answer ……………
How does this work?
3 of 48
0
Quadratic equations
If (x+a)(x+b) = 0 then either
x+a = 0 or x+b = 0
E.g. Solve: (3x – 15)(x + 7) = 0
As = 0 then either
(3x – 15) = 0 or (x + 7) = 0
Solve 3x – 15 = 0
and
+15 +15
3x
= 15
x
= 5
x+7= 0
–7 –7
x
= –7
to find solutions
Therefore either x = 5 or x = –7
4 of 48
Copy into
your notes
Quadratic equations
If (x+a)(x+b) = 0 then either
x+a = 0 or x+b = 0
E.g. Solve: (2x + 9)(x – 5) = 0
As = 0 then either
(2x + 9) = 0 or (x – 5) = 0
Solve 2x + 9 = 0
and
-9 -9
2x
= -9
x
= -4.5
x–5=0
+5 +5
x
=5
to find solutions
Therefore either x = -9 or x = 5
5 of 48
Your Turn
6 of 48
Question
Answer
(x + 3)(x – 5)=0
(x + 6)(x – 4)=0
(x – 7)(x + 5)=0
(2x + 8)(x – 6)=0
(2x – 7)(x + 9)=0
(2x + 9)(3x + 9)=0
(x – 8)(4x – 12)=0
-3 and 5
-6 and 4
7 and -5
-4 and 6
3.5 and -9
-4.5 and -3
8 and 3
Questions to do from the books
From factorised form
Achieve
Gamma P65 Ex5.01
CAT 1.2 P37 Q275–283
7 of 48
Merit
Excellence
SLO
Revision of quadratic factorisation
8 of 48
Revision of factorisation
Single brackets: i.e. 6x2 + 9x = 3x(2x + 3)
Quadratics (a = 1): i.e.
x2
x
+
+ 3x + 2 = (x + 1)(x + 2)
Difference of two squares: i.e. 9x2 – 16 = (3x + 4)(3x – 4)
Merit Quadratics (a ≠ 1): i.e. 2x2 – 5x – 12 = (2x + 3)(x – 4)
9 of 48
Questions to do from the books
Revision of Quadratic Factorisation
Achieve
P19 Ex2.07
Gamma
P21 Ex2.08
CAT 1.2
10 of 48
P32 Q224–244
P33 Q245–256
Merit
P22 Ex2.10
P35 Q263–274
Excellence
SLO
Solve quadratics (a = 1) by factorisation
http://www.youtube.com/watch?v=T5ORyW9G_8s
(Solve quadratic by factorize 1st)
11 of 48
Solving quadratic equations by factorization
Solve the equation x2 + 5x + 6 = 0
Step 1: Rearranging the equation so that all the terms are on
one side (already done!).
x2 + 5x + 6 = 0
Step 2: Factorize:
Find two integers that add to make 5 and multiply to make 6.
2 and 3 therefore (x + 2)(x + 3) = 0
Step 3: Solve both parts (make = 0)
x+2=0
x = -2
12 of 48
or
x+3=0
x = -3
Solving quadratic equations by factorization
Solve the equation x2 – 49 = 0
Step 1: Rearranging the equation so that all the terms are on
one side (already done!).
x2 – 49 = 0
Step 2: Factorize:
Notice that this is the difference of 2 squares.
(x + 7)(x – 7) = 0
Step 3: Solve both parts (make = 0)
x+7=0
x = -7
13 of 48
or
x–7=0
x=7
Solving quadratic equations by factorization
Solve the equation 5x2 + 15x = 0
Step 1: Rearranging the equation so that all the terms are on
one side (already done!).
5x2 + 15x = 0
Step 2: Factorize:
Take common factors out.
5x(x + 3) = 0
Step 3: Solve both parts (make = 0)
5x = 0
x=0
14 of 48
or
x+3=0
x = -3
Copy into
your notes
Solving quadratic equations by factorization
Solve the equation x2 – 5x = –4
Step 1: Rearranging the equation so that all the terms are on
one side (add 4 to both sides).
x2 – 5x + 4 = 0
Step 2: Factorize:
Find two integers that add to make –5 and multiply to make 4.
-1 and -4 therefore (x – 1)(x – 4) = 0
Step 3: Solve both parts (make = 0)
x–1=0
x=1
15 of 48
or
x–4=0
x=4
Solving quadratic equations by factorization
16 of 48
Your Turn:
Solve the following
Question
p  13 p  30 =0
2
a  5a  6 =0
m2  4m  12 =0
2
k  7k  30=0
2
w  3w  28 =0
t2 – 100 = 0
2g2 – g = 0
2
17 of 48
Answer
10 and 3
-6 and 1
6 and -2
-10 and 3
7 and -4
10 and -10
0 and 2
Questions to do from the books
factorise first a = 1
Achieve
Gamma P65 Ex5.02
CAT 1.2 P37 Q284 – 298
18 of 48
Merit
Excellence
SLO
How to use a Graphic Calculator
(you will not have this in the exam)
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Solve x2 + 5x + 6 = 0
1: Select EQUAtions
2: Select POLYtnomial
3: Select Degree 2 (as the highest power is 2)
4: using general form ax2 + bx + c = 0, input values of a, b and c
5: Select SOLVe
6: Two answers appear in
square brackets
Click for graphic calculator
20 of 48
20
SLO
Revision of quadratic factorisation (a ≠ 1)
http://www.youtube.com/watch?v=1T-rsltsWnM
( a  1 & no rearranging)
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E.g. 1) Factorise 3x2 + 17x + 20
x
2
3x
x
+ 17x + 20
Find two numbers that
multiply to give 3
The coefficients of x
must be 1 and 3
22 of 48
Find two numbers that
multiply to give 20
The constants could be
(1,20), (20,1), (2,10),
(10,2), (4,5), (5,4)
Continued on next slide
From previous slide
The coefficients of x
must be 1 and 3
The constants could be (1,20),
(20,1), (2,10), (10,2), (4,5), (5,4)
This leaves us with 6 possible answers. Only one is correct!
You can either expand each
one or use smiley face below.
(dx + e)(fx + g)
=(dg + ef)x
Or sum of inners and outers
give coefficient of x
Possible answer Yes or No
(1x + 1)(3x + 20)
No
(1x + 20)(3x + 1)
No
(1x + 2)(3x + 10)
No
(1x + 10)(3x + 2)
No
(1x + 4)(3x + 5)
Yes
(1x + 5)(3x + 4)
No
Therefore 3x2 + 17x + 20 = (x + 4)(3x + 5)
23 of 48
Factorizing quadratic expressions: Merit
24 of 48
Your Turn:
Factorise the following
Question
25t2 – 20t + 4
4y2 + 12y + 5
8t2 – 2t – 1
6x2 + 11x – 10
25 of 48
Answer
(5t – 2)(5t – 2)
(2y + 1)(2y + 5)
(4t + 1)(2t – 1)
(3x – 2)(2x + 5)
Questions to do from the books
Revision of quadratic factorisation
(a ≠ 1)
Achieve
Merit
Gamma P65 Ex5.02
P22 Ex. 2.10/11
CAT 1.2
P34 Q263–274
P37 Q284 – 298
Factorise
and solve
a=1
26 of 48
Excellence
SLO
Merit: Quadratic questions (a ≠ 1) and
rearrangement required
27 of 48
Solving quadratic equations by factorization
Solve the equation 6x2 = 31x – 35
Step 1: Rearranging the equation so that all the terms are on
one side.
6x2 – 31x + 35 = 0
Step 2: Factorize:
Product of 2 numbers = 6, product of another 2 numbers = 35
3x2 and -5x-7
(3x – 5)(2x – 7) = 0
Step 3: Solve both parts (make = 0)
3x – 5 = 0 or
5
𝑥=
3
28 of 48
2x – 7 = 0
7
𝑥=
2
Copy into
your notes
Solving quadratic equations by factorization
Solve the equation 8x2 + 2x = 15
Step 1: Rearranging the equation so that all the terms are on
one side (subtract 15 from both sides).
8x2 + 2x – 15 = 0
Step 2: Factorize:
Product of 2 numbers = 8, product of another 2 numbers = -15
2x4 and 3x-5 (2x + 3)(4x – 5) = 0
Step 3: Solve both parts (make = 0)
2x + 3 = 0 or
3
𝑥=−
2
29 of 48
4x – 5 = 0
5
𝑥=
4
Your Turn:
Solve the following
Question
25t2 – 20t + 4
4y2 + 12y + 5
2
8t – 2t – 1
2
6x + 11x – 10
30 of 48
Answer
2
2
and
5
5
1
5
− and −
2
2
1
1
− and
4
2
2
5
and −
3
2
Questions to do from the books
rearrange, factorise a ≠ 1
Achieve
Merit
Gamma P65 Ex5.02
P68 Ex5.03
CAT 1.2 P37 Q284 – 298
P38 Q299–310
Factorise
and solve
a=1
31 of 48
Excellence
Excellence examples
32 of 48
Excellence
Solve:
1
5
+
= 2
x
x+4
The first step when solving equations involving fractions is to
multiply through by the product of the denominators.
multiply by x(x + 4):
x + 4 + 5x = 2x(x + 4)
expand brackets and simplify:
6x + 4 = 2x2 + 8x
collect all terms on the r.h.s.:
0 = 2x2 + 2x – 4
divide by 2:
0 = x2 + x – 2
factorize:
0 = (x + 2)(x – 1)
x = –2
33 of 48
or
x=1
Excellence
Solve
4
3
–
= 1
x+2
x+8
Start by multiplying through by x + 2 and x + 8 to remove the
denominators,
4(x + 8) – 3(x + 2) = (x + 2)(x + 8)
expand the brackets:
4x + 32 – 3x – 6 = x2 + 10x + 16
x + 26 = x2 + 10x + 16
simplify:
collect all terms on the r.h.s.:
0 = x2 + 9x – 10
factorize:
0 = (x + 10)(x – 1)
x = –10
34 of 48
or
x=1
Excellence
Equations involving algebraic fractions may also lead to
quadratic equations that do not factorize. For example,
x
2
–
= 3
4–x
x
Solve,
Multiply through by x(4 – x):
expand the brackets:
collect terms on the l.h.s.:
divide by 2:
x2 + 2(4 – x) = 3x(4 – x)
x2 + 8 – 2x = 12x – 3x2
4x2 – 14x + 8 = 0
2x2 – 7x + 4 = 0
This quadratic equation cannot be solved by factorization so
we have to solve it using the quadratic formula.
35 of 48
Excellence Vocab
If 2 numbers follow each other they are consecutive.
The next consecutive number after x is x+1.
The next consecutive odd number after x is x+2.
36
36 of 48
Questions to do from the books
rearrange, factorise a ≠ 1
Achieve
Merit
Excellence
Gamma P65 Ex5.02
P68 Ex5.03
P68 Ex5.03
Q24, 25, 29, 30
CAT 1.2 P37 Q284 – 298
P38 Q299–310
Factorise
and solve
a=1
37 of 48
SLO
Worded quadratic questions
38 of 48
Find the width of the rectangle
The length of this rectangle is 4 cm more than its
width. The area of the rectangle is 45 cm 2.
Find the width of the rectangle.
If we call the width of the rectangle x we can draw the following
diagram:
x+4
x
Using the information about the area of the rectangle we can
write an equation:
x(x + 4) = 45
39 of 48
Quadratic equations
Multiply out the bracket,
x(x + 4) = 45
x2 + 4x = 45
Arrange quadratic equations so that all the terms are on one side.
x2 + 4x – 45 = 0
Factorise (click for hint)
Find two integers that add together to make 4 and multiply
together to make –45 (click for hint) i.e. 9 and –5
(x + 9)(x – 5) = 0
x = 5 and –9
x is a length so cannot be negative therefore x = 5 is the only
solution.
Solve to find x
40 of 48
Questions to do from the books
worded quadratic
Achieve
Gamma
CAT 1.2
41 of 48
Merit
P69 Ex5.04
Excellence
Random Achieve Revision questions
42 of 48
1 x  5x  6
2
6  6
x2  5x  6  0
( x  6)( x  1)  0
x  6, x  1
2
4 x  4 x  32
x 2  4 x  32  0
( x  8)( x  4)  0
x  8,
x  4
2
6 x  8x  0
x( x  8)  0
x  0 x 8  0
x  0, x  8
43 of 48
21) Solve
33) Solve
3 x(4 x  5)  0
(3 x  4)(5  7 x)  0
5
(3 x  4)  0, (5  7 x)  0
x  0, x 
4
3 x  4, 5  7 x
(2 x  7)(1  x)  0
4
5
x ,
x
7
3
7
x
, x 1
2
2
3
x
 9x  0
5
2
x  6 x  16  0
3 x( x  3)  0
( x  8)( x  2)  0
x  0, x  3
x  8, x  2
7 x 2  5 x  14  0
Extn : or calc
( x  7)( x  2)  0
2
6 x  17 x  10  0
x  7  0, x  2  0
(6 x  5)( x  2)  0
x  7 ,
x2
5
x , x2
6
1 4 x(3 x  2)  0
2
x  6x  0
2
x( x  6)  0
2
x  0, x 
3
4 ( x  3)( x  4)  0
2
x
 144  0
3
( x  12)( x  12)  0
x  0, x  6
x  3 x  70
2
x  3, x  4
x  12  0, x  12  0
x  12, x  12
x 2  3 x  70  0
( x  10)( x  7)  0
x  10, x  7
7 (3 x  2)(4 x  7)  0
3)
5 Solve
x  7x  0
2
x( x  7)  0
x  0, x  7
44 of 48
( x  3)2  0
( x  3)( x  3)  0
6
( x  3)  0
3
3
x3
either
or
3x  2  0
4x  7  0
3x  2
2
x ,
3
4 x  7
x
7
4
2
1 (3 x  5)(2 x  3)  0
(3 x  5)  0,(2 x  3)  0
3 x  5,
2 x  3
5
3
x ,
x
3
2
5
2
4 x  100  0
( x  10)( x  10)  0
x  10, x  10
7 x 2  x  20  0
( x  5)( x  4)  0
x  5, x  4
( x  7)( x  5)  0
Either
( x  5)  0
x7
x  5
x  8 x  16  0
2
( x  4)( x  4)  0
( x  4)  0
x4
x4  0
x  0, x  4
(3 x  7)(4 x  5)  0
either
or
( x  7)  0
8 x( x  4)  0
x  0,
3
6
(3 x  7)  0, (4 x  5)  0
7
7
5
5
3 x  7,
4 x  5
7
5
x ,
x
3
4
x 2  12 x  27  0
( x  9)( x  3)  0
x  9,
x  3
9 (4 x  5)(7 x  8)  0
either
or
(4 x  5)  0
5
5
(7 x  8)  0
8 8
7 x  8
4x  5
5
x
4
45 of 48
or
x
8
7
1
x  4x  0
2
x( x  4)  0
x  0,
x40
x  0,
x4
1) Solve
4
3 x(4 x  5)  0
5
x  0, x 
4
7 (2 x  7)(1  x)  0
7
x
, x 1
2
x 2  6 x  16  0
( x  8)( x  2)  0
x  8,
46 of 48
x  2
2 (4 x  1)(3 x  7)  0
either
or
(4 x  1)  0 (3 x  7)  0
4x  1
3 x  7
1
7
x ,
x
4
3
5 x 2  17 x  42  0
2
x
 3x  0
3
x( x  3)  0
x  0 x3  0
x0 x3
2
x
9  0
6
( x  3)( x  3)  0
( x  14)( x  3)  0
x  3  0, x  3  0
x  14, x  3
x  3, x  3
8 (2  3x)(7 x  4)  0
 either
or
(2  3x)  0
(7 x  4)  0
3x  3x
3x  2
2
x
3
4
4
7 x  4
4
x
7
Random Merit Revision questions
47 of 48
2
1 x  8 x  48
2 x 2  3 x  70
2
x
 x  72
3
x 2  8 x  48  0
x 2  3 x  70  0
x 2  x  72  0
( x  12)( x  4)  0
( x  10)( x  7)  0
( x  9)( x  8)  0
x  12, x  4
x  10, x  7
x  9,
x8
x 2  144  0
4 x 2  9 x  36
( x  12)( x  12)  0
x  12  0, x  12  0
x 2  9 x  36  0
x  12, x  12
( x  12)( x  3)  0
x  12, x  3
2)
5 Solve
x  4 x  32
2
x 2  9 x  36
x  4 x  32  0
x 2  9 x  36  0
( x  8)( x  4)  0
( x  12)( x  3)  0
x  8, x  4
x  12, x  3
2
48 of 48
6 Solve
3)
2
x
3  11x  42
x 2  11x  42  0
2 x2  5x  6
13) Find x
6  6
x2  5x  6  0
x4
( x  14)( x  3)  0
x  14, x  3
( x  6)( x  1)  0
A  56cm2
x3
either
or
x60
( x  4)( x  3)  56
x  6,
2
4 x  5x  6
x 1  0
x2  5x  6  0
( x  6)( x  1)  0
x  6, x  1
x 1
x 2  7 x  12  56
x 2  7 x  44  0
( x  11)( x  4)  0
2
6 x  3 x  70
x  11, x  4
Dimensions 7cm by 8cm
2
x
 4 x  32
5
x 2  4 x  32  0
( x  8)( x  4)  0
49 of 48
x  8,
x  4
x 2  3 x  70  0
( x  10)( x  7)  0
Either
or
( x  10)  0
( x  7)  0
x  10
x  7
Random Excellence Revision questions
50 of 48
Solve
( x  1) 5 x  x  8 ( x  1)
( x  3)2  25
x 1
( x  3)( x  3)  25
5x  ( x  8)( x  1)
x 2  3x  3x  9  25
5x  x2  x  8x  8
x 2  6 x  9  25  0
x2  7 x  5x  8  0
x  6 x  16  0
2
x2  2 x  8  0
( x  8)( x  2)  0
( x  4)( x  2)  0
x  8, x  2
x  4, x  2
5) Solve
5 x(7 x  2)  0
either 5 x  0 or 7 x  2  0
x0
51 of 48
2
x
7
A rectangle of sides 11cm by 16cm has the same amount added to
each side. The new area is 336cm 2 . Find how much was added to
each side.
16  x
Alw
11  x
(11  x)(16  x)  336
A  336cm2
176  16 x  11x  x 2  336
176  27 x  x 2  336
5cm added to each side
x 2  27 x  176  336  0
x  27 x  160  0
2
New dimensions 16cm by 21cm
( x  5)( x  32)  0
x  5,
52 of 48
x  32
can't have negative length
11) Solve
1)
2 Solve
3(2 x  1)  5
3x
25
7
6x  3  5
2
2
3
3
3x
7  77
6x  8
7
3 x  49
8
x
49
1
6
x   x  16
3
3
4
x

3x  2
3
75
 5 7
7
4x
5  7
3 x  2  35
6
3
2
2
3 x  37
37
x
3
1
x  12
3
53 of 48
+5
+5
1)
3 Solve
4(5 x  1)  3
1) Solve
4
20 x  4  3
5  6x  2  7
4
4
20 x  7
7
x
20
2x
1  7
7
3
+1 +1
4x
3   12  3
3
4 x  36
2x
3  8 3
3
2 x  24
x9
x  12
5  2(3 x  1)  7
3  6x  7
3
3
6x  4
4 2
x 
6 3
5x
8 4 27
+2 +2
5x
4  9 4
4
5 x  36
x
36
5
2)
2 Solve
4x
27
3
2 2
12) Solve
4(5 x  1)  3(2 x  7)  5
20 x  4  6 x  21  5
14 x  25  5
14 x  30
4x
9
3
4 x  27
15
x
7
5x 1
24
 3 2
2
5x 1  6
5x  7
7
x
5
27
x
4
2
6 ( x  3)  25
( x  3)( x  3)  25
x 2  3x  3x  9  25
x 2  6 x  9  25  0
5x  2
45
 74
4
5 x  2  28
2
5 x  30
x6
54 of 48
3 4(5 x  1)  16
20 x  4  16
4 4
20 x  20
x 1
2
x 2  6 x  16  0
( x  8)( x  2)  0
either
or
x  8  0,
x20
x  8, x  2