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Question 3 paper 1 1998.
p
 3,..t  0 .
5t
Here we are asked to write one letter in terms of another or to make p a subject of a
formula. Remember this is just an equation and you are looking for p on it’s own.
Solution
p
q   3  5tq  p  15t  p  15t  5tq . All we did here was multiply everything by
5t
the common denominator (5t) and bring p to one side.
(a) Express p in terms of q and t when q 
(B)(i) If (x – 2) is a factor of 3x 3  x 2  kx  6 find k. If (x – 2) is a factor then x = 2 is a
root if we sub in 2 for x we will get zero.
3(2) 3  (2) 2  k (2)  6  0  24  4  2k  6  0  2k  34  k  17 .
Be careful when you sub in numbers remember 3x 3  3( x)( x)( x) .
(B)(ii) Write down an equation, which has the three roots of value – 3,1,and 5.
Solution
We use the rule that if a is a root of an equation, x – a is a factor therefore the three
factors are (x - - 3), (x – 1) and (x – 5) to find the equation we multiply the factors.
( x  3)( x  1)( x  5)  ( x 2  x  3x  3)( x  5)  ( x 2  2 x  3)( x  5)  x 3  5 x 2  2 x 2  10 x  3x  15 
x 3  3x 2  13x  15
©(I) Write
1
2

as a single fraction.
x 1 x  3
Solution
This is just like adding any two fractions just get the common denominator and add! The
common denominator here is (x + 1)(x – 3).
1
2
1( x  3)  2( x  1) x  3  2 x  2
3x  1
(10 marks)




x 1 x  3
( x  1)( x  3)
( x  1)( x  3) ( x  1)( x  3)
(ii) Hence or otherwise solve
1
2

 1.
x 1 x  3
Solution
We can use our result from above (hence) or as we did in part a multiply everything by
the common denominator (otherwise).
We will this time use the result from part (i)
3x  1
 1  3x  1  1( x  1)( x  3)  3x  1  x 2  3x  x  3  0  x 2  5 x  2
( x  1)( x  3)
This is a quadratic equation, which has no factors so we must use the
 b  b 2  4ac
this gives
2a
 (5)  (5) 2  4(1)( 2)
5  25  8
5  33
(10 marks)
x
x
x
2(1)
2
2
formula x 
Leaving Cert 1997 QUESTION 3 PAPER 1
(a) Express p in terms of q and t when 2p – q = 3(p – t). Same as ’98 just keep your
eye on p.
Solution
2p – q = 3p –3t, get all the p’s to the same side 3t-q = 3p – 2p = 3t – q = p.
(Pray that there are lots of problems like this on the exam)
(b) Solve 2 x 3  3x 2  5 x  6  0 , this means find the values of x, which solve the
equation (Make the equation zero)
Solution
We find the first value by trial and error to get an idea as to what values of x to try first
divide –6 (end term) by the number in front of the x cubed
 6  2  3 , This means that 1, or –1 or 3 or –3 is a root (these are the factors of –3)
We use synthetic* division to see if – 1 is a root as can be seen the remainder is zero and
the quotient is 2 x 2  x  6
 1 2
3
56
2
1 6
1
60
We now solve 2 x 2  x  6  0 , by factors or the formula
2 x 2  x  6  0  (2 x  3)( x  2)  2 x  3  0, x  2  0  x  3 / 2, x  2
*Synthetic division is probably the best way to do this type of question as it avoids the
“dreaded” long division in algebra! Note one of the roots of the cubic must be a whole
number.
© Let f (x) = (2 + x)(3- x), Write down the solutions (roots) of f (x) = 0.
Solution
Just let (2 + x)(3 – x) = 0, 2 + x = 0, x = -2, 3 – x = 0, x = 3.
Let g (x) = 3x-k. The equation f (x) + g (x)=0has equal roots find one value for k
This was quite difficult for ordinary students, most would have no idea how to find k, ex
higher level students would probably have been able to find k, the fact that the question
was a wash out was reflected in the marking scheme where they gave 7 out of the 10
marks for writing out the question!
Solution
f ( x)  g ( x)  0  (2  x)(3  x)  3x  k  0(3marks)  6  2 x  3x  x 2  3x  k  0
 x 2  4 x  6  k  0(7marks)
If this has equal roots b 2  4ac  0  4 2  4(1)(6  k )  0  16  24  4k  0  k  10
Leaving Cert Ordinary Level 1999 Question 3 paper 1
p  3r
 5 . Same as ’98 and ‘97. Just
q
multiply everything by the common denominator (q) and get p on it’s own.
(a) Express p in terms of q and r when
Solution
p  3r
 5  p  3r  5q  p  5q  3r (10 marks).
q
x  2y  6
(b) Solve for x and y 2
a linear and a quadratic equation and we are
x  y 2  17
asked to solve them simultaneously. This type of problem comes up regularly in
the question on the circle. The standard method is as follows.
(1) Write the linear equation as x equals or y equals.
(2) Sub the result into the quadratic and simplify it.
(3) Solve the quadratic equation using factors or the formula.
(4) Substitute your answer(s) back into the line to get the values of the second
letter.
Solution
(1) x  2 y  6  x  6  2 y (2) substitute this result into the quadratic to get
(6  2 y) 2  y 2  17  (6  2 y)(6  2 y)  y 2  17  36  12 y  12 y  4 y 2  y 2  17
5 y 2  24 y  19  0  (5 y  19)( y  1)  0  5 y  19  0, y  1  0  y  4.8. y  1.
Now find x when y= 1 x = 6-2(1), x =4 our first result is (4,1)
When y = 4.8, x = 6 –2(4.8) x = 6 – 9.6 x –3.6 our second result is (-3.6,4.8)
©Show 6 x 2  5 x  4 is a factor of 6 x 3  17 x 2  6 x  8
Solution
Nothing for it but to divide it in
x2
6 x  5 x  4 )6 x  17 x 2
2
3
6 x  5x
3
12 x
2
2
 6x
8
 4x
 10 x  8
............................12 x 2 .......  10 x.....  8
................................0...............0..........0
Since the remainder is zero then 6 x 2  5 x  4 is a factor.
To find the three roots we set the two factors equal to zero
(6 x 2  5 x  4)( x  2)  0  (3x  4)( 2 x  1)( x  2)  0  3x  4  0  x  4 / 3, and .(2 x  1)  0  x  1
and ( x  2)  0  x  2
1996 Leaving cert Ordinary Level paper 1 Question 3
(a) Express q in terms of p and t when 2( p  3)  t .
Again this is just an equation we want q on it’s own this time.
Solution; 2( p  3q)  t  2 p  6q  t  2 p  t  6q 
2p t
 q (10 marks)
6
(b) Find the roots of the equation 2 x 3  5 x 2  x  2  0 .
Solution
Find the first root by trial and error to get an idea of what the first root might be divide 2
by 2 = 1 (last number divided by first number) Try x = 1 using Synthetic division.
1) 2  5  1  2
23 2
The remainder is zero therefore x = 1 is a root the quotient is
320
2 x 2  3 x  2 to find the other root set
2 x 2  3  2  0  (2 x  1)( x  2)  0  x  1 / 2, x  2
The three roots are –1/2,2,1
© Let f (x) = (1-x)(2+x) Write down the solutions of f (x) = 0.
Solution
(1  x)( 2  x)  0  1  x  0,2  x  0  1  x, x  2 .
(20 marks)
Find the range of values of x for which f (x) > 0.
This is a quadratic inequality (unusual for Ordinary level Maths)
To sole this we use the following method
(1) Solve f (x)=0, (2) Plot the results on the number line, plot zero also on the number
line.
(2) Sub zero into f (x) > 0 if true then zero is part of the solution, if false then zero is
not part of the solution.
(3) f ( x)  0  x  1, x  2
-2
0
1
(4) f(0)=(1)(2) > 0, therefore zero is part of the solution so our solution is
 2  x  1.
Leaving cert Ordinary level 1995 question 3 paper 1.
(a) Solve for x the equation 3 2 x  9 .
Solution:
This is an index equation, so change both sides to powers of the same number and set the
indices equal to each other.
32 x  9  32 x  32  2 x  2  x  1
(b) Solve for x and y.
x  2y 
3
x 2  y 2  26
Solution:
Simultaneous equations where one equation is linear, the second is a quadratic we use the
exact same method as in the’99 question.
(1) x  3  2 y,.....(2)(3  2 y) 2  y 2  26  (3  2 y)(3  2 y)  y 2  26  9  6 y  6 y  4 y 2  y 2  26
5 y 2  12 y  17  0  (5 y  17)( y  1)  0  y  17 / 5,....... y  1,  y  3.4
Now find x
We know x = 3 –2y, if y = - 1, x = 3 – (2)(-1) = 5 (5, -1)
Y = 3.4 x = 3 –2(3.4) = x = -3.8
(-3.8,3.4).
© f ( x)  x 2  bx  c , If f (x) = 0 when x = –3 and 1 find b and c
Solution
f (3)  (3) 2  b(3)  c  0  9  3b  c  0  3b  c  9
. f (1)  (1) 2  b(1)  c  0  1  b  c  0  b  c  1
These are simultaneous equations
 3b  c   9
bc 
1
Multiply the bottom line by – 1
b + c = - 1 , 2 + c = -1 c = - 3 , b =2 .
 3b  c   9
bc 
1
add –4b = -8, b = 2
f ( x)  x 2  2 x  3
If f (-1) = k find k.
Solution
f (1)  (1) 2  2(1)  3  k  4
Solve f (x) – k = 0. x 2  2 x  3  4  x 2  2 x  1  0  ( x  1)( x  1)  0  x  1.
Part c was really more awkward than difficult but it should have been possible to
get most of the question out if you read it carefully. The function notation is popular
in question 3 c.