(i) Factorise 3(x + 2)^2 - 10(x + 2) + 3
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We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:
- b = 1
- a = -1
- c = -2
- d = -1
<h3>Presenting the equation:</h3>
Basically, we want to solve:
The matrix product will be:
Then we must have:
-b + 2 = 1
This means that:
b = 2 - 1 = 1
- b = 1
We also need to have:
a*b + 1 = 0
we know the value of b, so we just have:
a*1 + b = 0
- a = -1
Now the two remaining equations are:
-c + 2d = 0
a*c + d = 1
Replacing the value of a we get:
-c + 2d = 0
-c + d = 1
Isolating c in the first equation we get:
c = 2d
Replacing that in the other equation we get:
-(2d) + d = 1
-d = 1
- d = -1
Then:
c = 2d = 2*(-1) = -2
- c = -2
So the values are:
- b = 1
- a = -1
- c = -2
- d = -1
If you want to learn more about systems of equations, you can read:
Given list of values
A = {23, 34, 19, 10, 9, 20, 40, 43, 28, 29, 36, 13, 27, 23, 36, 37, 7}
Sort list A to get list B
B = {7, 9, 10, 13, 19, 20, 23, 23, 27, 28, 29, 34, 36, 36, 37, 40, 43}
The smallest value is 7, so the min is 7
The largest value is 43, so the max is 43
Look through list B to find the middle-most value. What you can do is cross off each pair of outer values (see figure 1 in the attached images) until you are left with the middle value
Or you can note how there are 17 values so the median is at slot 9 (there are 8 values below the median; 8 values above the median; the 8 is from 17/2 = 8.5)
Once you know the median is 27, we split the list B into two parts
The first part, we'll call it L for lower half is
L = {7, 9, 10, 13, 19, 20, 23, 23}
which is the list of all values lower or smaller than the median.
The median of list L is 16 because its the midway point between 13 and 19. We can compute (13+19)/2 = 32/2 = 16. This is the value of Q1
The upper half U is
U = {28, 29, 34, 36, 36, 37, 40, 43}
this list shows all the values larger than the median.
The value of Q3 is 36. We find the midpoint of 36 and 36 which is just 36.
The value of Q3 is the same as the median of list U
------------------------------------------------
Five Number Summary:
Min = 7
Q1 = 16
Median = 27
Q3 = 36
Max = 43
This five number summary is then used to construct the box-and-whisker plot. See figure 2 in the attached images. The diagram shows how each piece of the boxplot is put together (eg: the left edge of the box is the value of Q1).
A = {23, 34, 19, 10, 9, 20, 40, 43, 28, 29, 36, 13, 27, 23, 36, 37, 7}
Sort list A to get list B
B = {7, 9, 10, 13, 19, 20, 23, 23, 27, 28, 29, 34, 36, 36, 37, 40, 43}
The smallest value is 7, so the min is 7
The largest value is 43, so the max is 43
Look through list B to find the middle-most value. What you can do is cross off each pair of outer values (see figure 1 in the attached images) until you are left with the middle value
Or you can note how there are 17 values so the median is at slot 9 (there are 8 values below the median; 8 values above the median; the 8 is from 17/2 = 8.5)
Once you know the median is 27, we split the list B into two parts
The first part, we'll call it L for lower half is
L = {7, 9, 10, 13, 19, 20, 23, 23}
which is the list of all values lower or smaller than the median.
The median of list L is 16 because its the midway point between 13 and 19. We can compute (13+19)/2 = 32/2 = 16. This is the value of Q1
The upper half U is
U = {28, 29, 34, 36, 36, 37, 40, 43}
this list shows all the values larger than the median.
The value of Q3 is 36. We find the midpoint of 36 and 36 which is just 36.
The value of Q3 is the same as the median of list U
------------------------------------------------
Five Number Summary:
Min = 7
Q1 = 16
Median = 27
Q3 = 36
Max = 43
This five number summary is then used to construct the box-and-whisker plot. See figure 2 in the attached images. The diagram shows how each piece of the boxplot is put together (eg: the left edge of the box is the value of Q1).