Download NEW ALGEBRA QUESTIONS

NEW ALGEBRA QUESTIONS Q11
Factorise ( x 2  1)
X
A:
( x  1) 2
X
B:
( x  1)2
Y
C:
( x  1)( x  1)
X
D:
Does not factorise!
X
E:
1 
1

 x   x  
2 
2

A:
Incorrect.
This multiplies out to x 2  2 x  1 .
The correct value is below:
B:
Incorrect.
This is equal to x 2  2 x  1 be careful
The correct value is below:
C:
Correct.
A nice and easy one?
The correct value is below, you don't need to look, do you?:
D:
Incorrect.
Occasionally this will be the correct answer. Did you even try?
The correct value is below:
E:
Incorrect.
½ times ½ equals ¼ .
This is the difference of two squares ( x2  1)  ( x  1)( x  1)
Q12
Factorise ( x 2  2 x  1)
Y
A:
( x  1) 2
X
B:
( x  1)2
X
C:
( x  1)( x  1)
X
D:
Does not factorise!
X
E:
1 
1

 x   x  
2 
2

A:
Correct.
They will get harder! It's all a random chance!
The correct work is below if you need it.
B:
Incorrect.
This will give a -2x in the middle. Be careful.
The correct work is below if you need it.
C:
Incorrect.
This is the difference of two squares and the middle value vanishes.
The correct work is below if you need it.
D:
Incorrect.
Or is it? Look below for the answer.
The correct work is below if you need it.
E:
Incorrect.
Only the x2 term will be correct. You could not have done much worse.
The correct work is below and you need it.
The factors are ( x  1) and ( x  1)  ( x  1)2
Q13
Find the roots of ( x2  2 x  3)
X
A:
3,1
X
B:
1,3
X
C:
1,-3
X
D:
The roots are not real.
Y
E:
3,-1
A:
Incorrect.
Funny, having two answers the same. Guess they are both wrong.
The correct work is shown below:
B:
Incorrect.
You should be suspicious. This is exactly the same answer as in A, so expect it to
be wrong! The correct work is shown below:
C:
Incorrect.
Equate the expression to zero and factorise it. You have got the signs wrong at the
last line. Bad luck. The correct work is shown below:
D:
Incorrect.
For this to be the case the roots would have to involve the root of a negative.
That is part of complex number theory and will come later. Oh yes it will!
The correct work is shown below:
E:
Correct.
It was'nt really that difficult. Was that your first or fifth try?
The correct work is shown below:
Remember that roots are the values of the unknown that make the expression equal to zero.
i.e. ( x2  2 x  3) =0 (factorise) gives ( x  1)( x  3)  0 therefore ( x  1)  0 or ( x  3)  0
Therefore x  3,  1 are the two roots.
Alternatively you can solve the quadratic equation with the so called formula method or do it by trial and
error, trying x= 0, 1,2,3 etc. I know which I would do.
Q14
Given that x =-1 is a root of x3  5 x 2  8 x  4  0 find the other two. You need to do some long division or
solve by trying successive values of x. Then we could be here all night so go the long division route
(pun).
X
A:
+2,-2
X
B:
+1,-2
X
C:
+2,+2
Y
D:
-2,-2
X
E:
-1,+2
A:
Incorrect.
After the long division stage, these roots would give you the difference of two
squares. The correct work is shown below:
B:
Incorrect.
Your long division is probably wrong. The correct work is shown below:
C:
Incorrect.
You have got a double root but the sign is incorrect. A mistake near the end
perhaps? The correct work is shown below:
D:
Correct.
Well done or was it a lucky guess? Can you really do algebraic long division?
The full work is shown below:
E:
Incorrect.
The minus 1 would mean of the three roots there is a double on at -1. Check your
work carefully. The correct work is shown below:
x2  4x  4
x  1) x 3  5 x 2  8 x  4
x3  x 2
4x2  8x
Now factorise x 2  4 x  4 to give ( x  2)( x  2)
4x2  4x
4x  4
4x  4
Therefore the other roots are -2 and -2 (a double root!)
Q15
Split into partial fractions:
2x
x 1
2
Y
A:
1
1

x 1 x 1
X
B:
1
1

x 1 x 1
X
C:
2
1

x 1 x 1
X
D:
2
2

x 1 x 1
X
E:
1
1

x 1 x 1
A:
Correct.
It is a simple problem and you can almost see the answer but you have to be careful
with the signs. The correct work is shown below:
B:
Incorrect.
It is a simple problem and you can almost see the answer but you have to be careful
with the signs. Check through carefully. The correct work is shown below:
C:
Incorrect.
It is a simple problem and you can almost see the answer but you have to be careful
with the signs. Check through carefully. The correct work is shown below:
D:
Incorrect.
It is a simple problem and you can almost see the answer but you have to be careful
with the signs. Check through carefully. The correct work is shown below:
E:
Incorrect.
It is a simple problem and you can almost see the answer but you have to be careful
with the signs. Check through carefully. The correct work is shown below:
2x
2x
A
B
A( x  1)  B( x  1)
2x




 2
( x  1)( x  1)
x  1 ( x  1)( x  1) x  1 x  1
x 1
 A( x  1)  B( x  1)  2 x which is true for all x
2
*substitute x  1 gives A  1, substitute x  1 gives B  1
2x
1
1
 2


x 1 x 1 x 1
Now check the answer by cross multiplying.
* An alternative method from this point, is to equate the coefficients of
like powers of x and solve the ensuing simultaneous equations
Q16
Split into partial fractions:
x2  2 x  1
x2  1
X
A:
1
1

x 1 x 1
X
B:
1
2x
x 1
Y
C:
1
1
1

x  1 x 1
X
D:
1
1
1

x 1 x 1
X
E:
1
1
1

x 1 x 1
A:
Incorrect.
You have to use long division first to obtain a fraction with the numerator
of lower order than the denominator. The correct work is shown below:
B:
Incorrect.
Ok, you have done the long division. Now find the partial fractions.
The correct work is shown below
C:
Correct.
Well done. Just check that it was'nt luck.
The correct work is shown below:
D:
Incorrect.
The long division is OK but the partial fractions are wrong.
The correct work is shown below:
E:
Incorrect.
The long division is OK but the partial fractions are wrong.
The correct work is shown below:
2
First use long division to make the numerator of lower order than the denominator
1
x  1) x 2  2 x  1
2
1
x2
2x

x2  2 x  1
2x
1 2
2
x 1
x 1
Now find the partial fractions:
2x
2x
A
B
A( x  1)  B ( x  1)
2x




 2
(
x

1)(
x

1)
x

1
x

1
(
x

1)(
x

1)
x 1
x 1
 A( x  1)  B ( x  1)  2 x which is true for all x
2
substitute x  1 gives A  1, substitute x  1 gives B  1 
This makes the final answer
x2  2x  1
1
1
1

2
x

1
x
1
x 1
2x
1
1


x 1 x 1 x 1
2
Q17
Split into partial fractions:
1
x  x  x 1
3
2
.
Hint: you must find the factors of the denominator first, by what ever means
X
A:
1
1
1


( x  1) ( x  1) ( x  1)2
X
B:
1
1
 2
( x  1) ( x  2 x  1)
X
C:
1
1

( x  1) ( x  1)
Y
D:
1
1
1


4( x  1) 4( x  1) 2( x  1)2
Y
E:
1
x3

2
4( x  1) 4( x  2 x  1)
A:
Incorrect.
You have the correct form of partial fractions (with repeated factors).
However the constants are all wrong. See below for one way of doing it.
B:
Incorrect.
You have chosen another approach, using a quadratic expression rather than
repeated factors, but your numbers are wrong. See below for one way of doing it.
C:
Incorrect.
This way might have worked but you have made a mistake somewhere.
See below for one way of doing it.
D:
Correct.
This is the easiest way to do it, see below, but answer E is an alternative:
E:
Correct.
Not perhaps the easiest way but you have a correct answer. The given solution
uses the repeated factor method:
2
1
x  x  x 1
3
2

1
1
A
B
C




( x  1)( x  1)( x  1) ( x  1)( x  1) 2 ( x  1) ( x  1) ( x  1) 2
 A( x  1) 2  B( x  1)( x  1)  C ( x  1)  1
1
gives C   ,
2
2
equate coefficients of x gives A  B  1  A   B
substitute x  1
equate coefficients of x 0 (i.e. pure numbers) gives A  B  C  1  A  B 
1
1
1
Combining gives A  , B   , C  
4
4
2
1
1
1
1



x3  x 2  x  1 4( x  1) 4( x  1) 2( x  1) 2
NOW CHECK IT BY MULTIPLYING IT OUT!
1
2
Q18
Find the x values for the points of intersection of the following two functions: y1  x 2  4 and y2  12
X
A:
0
Y
B:
4
X
C:
4
X
D:
2
X
E:
16
A:
Incorrect.
This would mean both functions go through the origin, and in fact neither do.
The correct work is given below:
B:
Correct.
Equate the value of y and solve for x.
The correct work is given below:
C:
Incorrect.
There are two points given by the roots of 16.
The correct work is given below:
D:
Incorrect.
You should arrive at x2  16
The correct work is given below:
E:
Incorrect.
You need to take the square root of this.
The correct work is given below:
y1  x 2  4 and y2  12
put y1  y2  x 2  4  12 solve for x, x  4
Sketch it to be sure.
Q19
Solve the following simultaneous equations: x  y  3, 3x  y  1
Y
A:
x=1, y=2
X
B:
x=y=1
X
C:
x=y=2
X
D:
x=2, y=1
X
E:
x=3, y=0
A:
Correct.
This problem is so simple that you can probably see the answer. However
check through the method given below:
B:
Incorrect.
You really should not have got this wrong. This problem is so simple that you can
probably see the answer. However check through the method given below:
C:
Incorrect.
Surely you can see that this is wrong! This problem is so simple that you can
probably see the answer. However check through the method given below:
D:
Incorrect.
Close. You have got the numbers round the wrong way.
Check through the method given below:
E:
Incorrect.
Ok, this satisfies the first equation as do an infinity of answers!
Check through the correct method given below:
Eliminate one of the unknowns between the two equations.
Simplest is to add them and then y cancels.
(otherwise you have to multiply the equations to obtain the same result)
x  y  3 (i )
3x  y  1 (ii )
4x
add
 4  x  1 and y  2
Q20
Solve the following simultaneous equations: 3x  2 y  12, 4 x  y  11
X
A:
x  4, y  0
X
B
x  3, y  2
X
C:
They cannot be solved.:
X
D:
There is an infinite numbers of solutions.
Y
E:
x  2, y  3
A:
Incorrect.
You need to be a bit more careful, those values will not satisfy the second
equation. Check through the steps below:
B:
Incorrect.
You have the values the wrong way round. Is that just bad luck or a bad attempt at
guessing? Check through the steps below:
C:
Incorrect.
Sometimes this is true but not this time Moriarty!
Check through the steps below:
D:
Incorrect.
OK, so you think you know it all You are wrong in this case. (again?)
Check through the steps below:
E:
Correct.
Was that a lucky guess or just an insight? Either way, check through the steps in
the solution below:
3x  2 y  12 (i)
4 x  y  11 (ii)
3x  2 y  12
(ii)  2 and subtract
8 x  2 y  22
 5x
 10  x  2, y  11  8  3
Q21
Combine the following expressions into the simplest single fraction:
2
3
 2
x  2 x  4x  4
X
A:
7
x  4x  4
X
B
2x  4
x  4x  4
X
C:
2 x 2  11x  14
x3  6 x 2  12 x  8
X
D:
Y
E:
A:
Incorrect.
You have lost an x term. The correct work is shown below:
B:
Incorrect.
You have forgotten the 3. The correct work is shown below:
C:
Incorrect.
OK this is correct so far but it is not the simplest. Use factors as shown below:
D:
Incorrect.
You have lost the first term completely. Take more care.
The correct work is shown below:
E:
Correct.
Factorising the quadratic certainly helps! See below:
2
2
3
 x  2  x  2 
2x  7
x  4x  4
2
Note, the quadratic term factorises and also is the common denominator for the addition
2
3
2
3
2( x  2)  3
2x  7
 2



 2
x  2 x  4 x  4 x  2  x  2  x  2   x  2  x  2  x  4 x  4
Q22
Combine the following expressions into the simplest single fraction:
3
4x
 2
x 1 x 1
Y
A:
7x  3
x2  1
X
B
3x  1
x2  1
X
C:
X
D:
4x
x 1
X
E:
7x  3
x2  1
A:
Correct.
Factorising the quadratic certainly helps! See below:
B:
Incorrect.
You have forgotten the x in 4x. The correct work is shown below:
C:
Incorrect.
OK this is correct so far but it is not the simplest. Use factors as shown below:
D:
Incorrect.
You have somehow lost the first term completely. Take more care.
The correct work is shown below:
E:
Incorrect.
I think you have put x-1 on the denominator of the first tem by mistake.
Take more care. The correct work is shown below:
7 x2  4 x  3
 x  12  x  1
2
Note, the quadratic term factorises as the difference of two squares
and also is the common denominator for the addition
3
4x
3
4x
3( x  1)  4 x 7 x  3
 2




x  1 x  1 x  1  x  1 x  1  x  1 x  1 x 2  1
Q23
Solve the following equation for y: y 
4  x2
y  2x
X
A:
y=2
X
B
y=-2
X
C:
x
X
D:
x
Y
E:
y x2
A:
Incorrect.
Clearly x figures in the solution. See below for the correct work:
B:
Incorrect.
Clearly x figures in the solution. See below for the correct work:
C:
Incorrect.
You can see this will not work and in any case it is a quadratic in y, so you need 2
answers. See below for the correct work:
D:
Incorrect.
This looks closer but what about the 4! See below for the correct work:
E:
Correct.
Just go through the motions and it all comes out.
See below for the correct work:
y
4  x2
cross multiply y 2  2 yx  4  x 2 rearrange y 2  2 yx  x 2  4  0
y  2x
use formula: y 
2 x  4 x 2  4( x 2  4)
 x2
2
Q24
Find the points where the given function cuts the x and y axes x2  y 2  4 and deduce the shape of the
function
X
A:

X
B
(0,0)
Y
C:
(0,2) (2,0) (0,-2) (-2,0)
X
D:
(0,2) (2,0)
X
E:
(0,-2) (-2,0)
A:
Incorrect.
OK this is a point on the function but not where it crosses an axis.
See correct work below:
B:
Incorrect.
This is the origin and the function never goes near it!
See correct work below:
C:
Correct.
Right answers! Put a circle round it. See correct work below if you need to:
D:
Incorrect.
This is only half the answer. There are 4 points. See correct work below:
E:
Incorrect.
This is half of the answer and D is the other half.
There are 4 points. See correct work below:
2, 2

x2  y 2  4
put y  0, then x  2, similarly put x  0, then y  0
The function is that of a circle centred on the origin and of radius 2.
Q25
Find the 3 roots of the equation: x3  2 x 2  x  2 =0 either by trial, or by trial and the formula for the roots
of a quadratic, or by trial and factorising.
X
A:
+1,+2,-2
X
B:
+1,-2,-2
X
C:
-2,-2,-2
Y
D:
+1,-1,-2
X
E:
-1,+1,+2
A:
Incorrect.
With two roots of value 2 you would expect a 4 in the equation. However you have
found two correct roots. The correct work is shown below:
B:
Incorrect.
A double root of -2 would possibly indicate a 4 in the equation.
The correct work is shown below:
C:
Incorrect.
This looks very unlikely since ( x  2)3  x3  6 x2  12 x  8 which you should
know! The correct work is shown below:
D:
Correct.
Well done or was it a lucky guess? Can you really do algebraic long division?
The full work is shown below:
E:
Incorrect.
This looks close but the pure number at the end would be +2 not -2.
The correct work is shown below:
x3  2 x 2  x  2, by trial try x  1, 2, 3. However x  1 is easily found as a root.
By long division the quadratic left when x3  2 x 2  x  2 is divided by ( x  1) is:
x 2  3x  2  ( x  1)( x  2). The roots are therefore x  1, 1, 2.
In this case the three roots can be found quite quickly by trial, but this is not normally the case.
Q26
Expand: ( x  a)4
X
A:
x 4  x3 a  x 2 a 2  xa 3  a 4
X
B
x3  x 2 a  xa 2  a 3
X
C:
x3  3x 2 a  3xa 2  a3
Y
D:
x 4  4 x3 a  6 x 2 a 2  4 xa3  a 4
X
E:
x 4  2 x3 a  3x 2 a 2  4 xa3  a 4
A:
Incorrect.
You are missing the coefficients of each term. See correct work below:
B:
Incorrect.
You have expanded ( x  a)3 and still left out the coefficients.
See correct work below:
C:
Incorrect.
You have expanded ( x  a)3 but have got the coefficients wrong.
See correct work below:
D:
Correct.
Well done. Note the way the coefficients are symmetrical about the centre.
See correct work below:
E:
Incorrect.
The coefficients are all wrong and it looks like you have been guessing again!.
Practice this form of expansion, since it is very useful. See correct work below:
( x  a)4  x 4  4 x3a  6 x 2 a 2  4 xa 3  a 4 How?
Either by ( x 2  2ax  a 2 )2
or by a useful method you may know called the binomial expansion.
Q27
How many ways can you arrange 6 different objects in a row?
Y
A:
720
X
B:
360
X
C:
120
X
D:
6
X
E:
1
A:
Correct.
Yes. Interestingly the last little piggy has no choice where he goes. Not a
wonder he cried all the way home! See below for correct work:
B:
Incorrect.
This is the number of permutations of only 4 of the 6. Interestingly enough, 5 from
from 6 gives you the same answer as all 6 from 6. I feel sorry for the last one,
nobody cares about him! The correct work is given below:
C:
Incorrect.
This is actually 5! and happens to be the correct answer if you arrange them in a
circle, since there is no beginning or end and now it makes no difference where the
first one goes! SPOOKEY. The correct work is given below:
D:
Incorrect.
You are not trying now! They are all different and the order matters so juggle them
up a bit. The correct work is given below:
E:
Incorrect.
Funnily enough this would be correct if they were all the same.
The correct work is given below:
Yes, the first one can go in 6 places then the second in 5 etc. etc. down to 1, making 720.
This is known as the permutation of 6 item selected from 6, expressed mathematically:
6
P6 
6!
 6  5  4  3  2 1  720
0!
Remember the general expression n Pr 
Did I mention 0!=1 (weired?)
n!
(n  r )!
Q28
How many permutations (order matters) are there when you select 4 objects from 6 different ones?
X
A:
720
Y
B:
360
X
C:
120
X
D:
24
X
E:
15
A:
Incorrect.
This is the number of permutations of all 6. The correct work is given below:
B:
Correct.
We are only selecting 4 from the 6. The correct work is given below:
C:
Incorrect.
This is the answer selecting 4 from only 5, i.e. 5 P4 
5!
. Be more careful!
1!
The correct work is given below:
D:
Incorrect.
This is the number of ways of permuting 4 amongst themselves. However you can
select the 4 from the 6 in different ways. (actually in 30 ways, but that is another
problem). The correct work is given below:
E:
Incorrect.
This is the number of combinations of 4 selected from 6 where the order does not
matter. Each of the 4 can then be juggled in 4! (= 24) ways to give the total answer
of 360. The correct work is given below:
Yes, the first one can go in 6 places then the second in 5 the next in 4 and the last in 3
This is known as the permutation of 4 item selected from 6, expressed mathematically:
6
P4 
6!
6  5  4  3  2 1

 360
(6  4)!
2 1
Remember the general expression n Pr 
n!
(n  r )!
Q29
How many combinations (order does not matters) are there when you select 4 objects from 6 different
ones?
X
A:
720
X
B:
360
Y
C:
15
X
D:
5
X
E:
1
A:
Incorrect.
This is the number of permutations of all 6. The correct work is given below:
B:
Incorrect.
You have used the formula for permutations not for combinations.
The correct work is given below:
C:
Correct.
Well done. It is a relatively small number!
The correct work is given below:
D:
Incorrect.
You have worked out the number of combinations of 4 selected from 5 by mistake.
The correct work is given below:
E:
Incorrect.
You have worked out the number of combinations of all 6.
The correct work is given below:
C4 
6
6!
6  5  4  3  2 1

 15
(6  4)!4! 2 1 4  3  2 1
Remember the general expression nCr 
n!
(n  r )!r !
You would expect the reduction by a factor r! over the number of permutations, since each r selected can
be arranged amongst themselves in r! ways.
Q30
Use original question 7 from Basics