Cavendish University

SCHOOL OF MEDICINE

MSF Programme

MTM 120A:Foundation Mathematics

May 15, 2025

CAT 2 (Marking Key) Lecturer: Mr. PC Nkandu.

1
 SECTION A(35 marks)

(1) By completing the square

√
method, show that√the roots of the equation ax2 +bx+c = 0
b2 −4ac 2 −4ac
are given by x = −b + 2a or x = −b − b 2a . Hence, use the solution to solve
2
2x + x − 6 = 0.

Solution

Start with the general quadratic:

ax2 + bx + c = 0

Divide both sides by a (assuming a 6= 0):

b c
x2 + x + = 0
a a
b c
=⇒ x2 + x = −
a a
 2  2
2 b b c b
=⇒ x + x + =− +
a 2a a 2a
2
b2

b c
=⇒ x + =− + 2
2a a 4a
2
−4ac + b2

b
=⇒ x + =
2a 4a2
√
b b2 − 4ac
=⇒ x + =±
2a 2a
Solve for x: √
b b2 − 4ac
x=− ±
2a 2a
which simplifies to the quadratic formula:
√
−b ± b2 − 4ac
x=
2a
Final answer
√
−b ± b2 − 4ac
x=
2a

Application: Solve 2x2 + x − 6 = 0

Here, a = 2, b = 1, c = −6.
Applying the quadratic formula:
√ p
−b ± b2 − 4ac −1 ± 12 − 4 × 2 × (−6)
x= =
2a 2×2

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 Calculate the discriminant:

b2 − 4ac = 1 − 4 × 2 × (−6) = 1 + 48 = 49
√ −1±7
Square root: 49 = 7 Substitute: x = 4
Find the roots:

−1 + 7 6 3
x= = =
4 4 2
−1 − 7 −8
x= = = −2
4 4

Final answer:

3
the roots are x = and x = −2 .
2

(2) Given that (3x + 1) is a factor of f (x) = ax4 − x3 − bx2 + x + 1 and when divided by
(x − 2) the remainder is 63. Find the value of a and b. Hence, factorise the polynomial
completely.

Solution

Given the polynomial: f (x) = ax4 − x3 − bx2 + x + 1 and the conditions:

• (3x + 1) is a factor of f (x),

• When divided by (x − 2), the remainder is 63.

Since (3x + 1) is a factor, then:  

1
f − =0
3

Calculate f − 13 :

   4  3  2  
1 1 1 1 1
f − =a − − − −b − + − +1=0
3 3 3 3 3
a 1 b 1
=⇒ + − − +1=0
81 27 9 3
a + 3 − 9b − 27 + 81
=⇒ =0
81
=⇒ a − 9b + 57 = 0
∴ a − 9b = −57.........................(1)

The remainder when dividing by x − 2 is f (2) = 63:

f (2) = a(2)4 − (2)3 − b(2)2 + 2 + 1

=⇒ 16a − 8 − 4b + 2 + 1 = 63

Page 3 of 14
 =⇒ 16a − 4b − 5 = 63
=⇒ 16a − 4b = 68
∴ 4a − b = 17...........................(2)

Solving equation (2)

b = 4a − 17

Solve the simultaneous equations. Substitute b into Equation (1):

=⇒ a − 9(4a − 17) = −57

=⇒ a − 36a + 153 = −57

=⇒ −35a + 153 = −57
=⇒ −35a = −210
−210
a= =6
−35
b = 4a − 17 = 4 × 6 − 17 = 24 − 17 = 7

Final answer:

a = 6, b=7

Factorising the polynomial completely:

6x4 − x3 − 7x2 + x + 1
Divide , by applying the long division method we get,
3x + 1

f (x) = (3x + 1)(2x3 − x2 − 2x + 1)

Factor the cubic 2x3 − x2 − 2x + 1

Attempt rational roots: factors of 1 over factors of 2: ±1, ± 12
Test x = 1:
2(1)3 − (1)2 − 2(1) + 1 = 2 − 1 − 2 + 1 = 0
Yes, x = 1 is a root.
Divide by (x − 1) using synthetic division: The remaining quadratic is:

2x2 + x − 1

Factor the quadratic

2x2 + x − 1

Calculate discriminant

∆ = 12 − 4 × 2 × (−1) = 1 + 8 = 9

Page 4 of 14
 √
9=3

Find roots: √
−b ± ∆ −1 ± 3
x= =
2a 2×2
−1 + 3 2 1
x= = =
4 4 2
−1 − 3 −4
x= = = −1
4 4
Final factorization: f (x) = (3x + 1)(x − 1)(2x + 1)(x + 1)

Final answer

f (x) = (3x + 1)(x − 1)(2x + 1)(x + 1)

(3) Given α and β are the roots of x2 + 7x + 10 = 0, find

(a) α2 + β 2

Solution

Given the quadratic:

x2 + 7x + 10 = 0
By using Vieta’s formulas
b 7
α+β =− = − = −7
a 1
c 10
αβ = = = 10
a 1

We now find α2 + β 2 :
Recall the identity:
α2 + β 2 = (α + β)2 − 2αβ
Substitute the known values:

α2 + β 2 = (−7)2 − 2 × 10 = 49 − 20 = 29

Final answer

α2 + β 2 = 29

α β
(b) β + α

Solution

Page 5 of 14
 We need to find:
α β
+
β α
This can be written as:
α2 + β 2
αβ
Recall that:
α2 + β 2 = (α + β)2 − 2αβ

Substituting the known values:

α2 + β 2 = (−7)2 − 2 × 10 = 49 − 20 = 29

Thus:
α β 29
+ =
β α 10

Final Answer

α β 29
+ =
β α 10

(c) α − β

Solution

Recall:
(α − β)2 = (α + β)2 − 4αβ

Substitute:
(α − β)2 = 49 − 4 × 10 = 49 − 40 = 9
√
∴ α − β = ± 9 = ±3
Final Answer

α − β = ±3

(4) Solve the equation

y = 4x2 + 3ax − 2a2

y = 2x2 + 2ax − a2

Solution:

Given:
y = 4x2 + 3ax − 2a2 and y = 2x2 + 2ax − a2

Page 6 of 14
 Set the equations equal

4x2 + 3ax − 2a2 = 2x2 + 2ax − a2

Bring all to one side

(4x2 − 2x2 ) + (3ax − 2ax) + (−2a2 + a2 ) = 0

2x2 + ax − a2 = 0

To find the solution we use the quadratic formula where a = 2, b = a, and c = −a2 :
√
−b ± b2 − 4ac
x=
2a
p
−a ± a2 − 4(2)(−a2 )
x=
2×2
√
−a ± a2 + 8a2
x=
4
√
−a ± 9a2
x=
4
−a ± 3a
x=
4
Roots,
−a + 3a 2a a
x= = =
4 4 2
−a − 3a −4a
x= = = −a
4 4

Final answer:

a
x= or x = −a
2
[9,9,8,9]

SECTION B(45 marks)

QUESTION 1

(a) Given that x2 + px + q and 3x2 + q have a common factor (x − b), where p, q and b
are non zeros, show that 3p2 + 4q = 0.

Solution:
Since both x2 + px + q and 3x2 + q share the factor (x − b), then:

b2 + pb + q = 0 and 3b2 + q = 0

Page 7 of 14
 From the second equation:
3b2 + q = 0 ⇒ q = −3b2

Substituting q = −3b2 into the first:

b2 + pb − 3b2 = 0 ⇒ −2b2 + pb = 0

Factor out b (since b 6= 0):

b(−2b + p) = 0 ⇒ −2b + p = 0 ⇒ p = 2b
Calculate 3p2 + 4q:
3(2b)2 + 4(−3b2 ) = 3(4b2 ) − 12b2 = 12b2 − 12b2 = 0

Final Answer

3p2 + 4q = 0 .

(b) Decompose
3x+7
(i) (x+2)(x+3)

Solution
Express as partial fractions:
3x + 7 A B
= +
(x + 2)(x + 3) x+2 x+3
Multiply both sides by (x + 2)(x + 3):
3x + 7 = A(x + 3) + B(x + 2)

Expand:
3x + 7 = Ax + 3A + Bx + 2B = (A + B)x + (3A + 2B)

Matching coefficients: (
A+B =3
3A + 2B = 7
Solve:
From the first: B = 3 − A
Substitute into the second:
3A + 2(3 − A) = 73A + 6 − 2A = 7A + 6 = 7A = 1

Then,
B =3−1=2

Final answer:

3x + 7 1 2
= +
(x + 2)(x + 3) x+2 x+3

Page 8 of 14
 2x+7
(ii) (x−3)2

Solution

Express as partial fractions:

2x + 7 A B
= +
(x − 3)2 x − 3 (x − 3)2

Multiply both sides by (x − 3)2 :

2x + 7 = A(x − 3) + B

Expand:
2x + 7 = Ax − 3A + B

Matching coefficients: (
A=2
−3A + B = 7

Substitute A = 2:

−3(2) + B = 7
−6 + B = 7
B = 13

Final answer:

2x + 7 2 13
2
= +
(x − 3) x − 3 (x − 3)2

[5,5,5]

QUESTION 2
x3 +2x−5 r
(a) Given that x−1 ≡ ax2 + bx + c + x−1 , find the value of a, b, c and r.

Solution:

Dividing x3 + 2x − 5 by x − 1 using synthetic division we first Set up synthetic

division:
1 1 0 2 −5
1 1 3
1 1 3 −2

• The quotient is x2 + x + 3.
• The remainder is −2.

Page 9 of 14
 Therefore,
x3 + 2x − 5 = (x − 1)(x2 + x + 3) − 2,

which implies
x3 + 2x − 5 −2
= x2 + x + 3 + .
x−1 x−1
Final answer

a = 1, b = 1, c = 3, r = −2

(b) Solve |2x − 3| < 3.

Solution:

|2x − 3| < 3
⇒ −3 < 2x − 3 < 3 (by definition of absolute value)
Add 3 to all parts: − 3 + 3 < 2x − 3 + 3 < 3 + 3
0 < 2x < 6
0 6
Divide all parts by 2: <x<
2 2
0<x<3

Final answer:

x ∈ (0, 3)
m−2 2m
(c) Solve 3 +1= 7 .

Solution:
m−2 2m
To solve +1=
3 7
Multiply through by the least common denominator, 21:

 
m−2 2m
21 × + 21 × 1 = 21 ×
3 7
7(m − 2) + 21 = 3 × 2m
7m − 14 + 21 = 6m
7m + 7 = 6m
7m − 6m = −7
m = −7

Final answer:

m = −7

Page 10 of 14
[8,3,4]

QUESTION 3
x3 +y 3 x2 +xy+y 2
(a) Simplify x3 −y 3
× (x−y)2 +xy
.

Solution:

Factor numerator and denominator of the first fraction:

x3 + y 3 = (x + y)(x2 − xy + y 2 )
x3 − y 3 = (x − y)(x2 + xy + y 2 )

So,
x3 + y 3 (x + y)(x2 − xy + y 2 )
=
x3 − y 3 (x − y)(x2 + xy + y 2 )

Recall:
(x − y)2 + xy = (x2 − 2xy + y 2 ) + xy = x2 − xy + y 2

Thus,
x2 + xy + y 2 x2 + xy + y 2
=
(x − y)2 + xy x2 − xy + y 2

(x + y)(x2 − xy + y 2 ) x2 + xy + y 2
= ×
(x − y)(x2 + xy + y 2 ) x2 − xy + y 2

Notice that x2 + xy + y 2 and x2 − xy + y 2 cancel out:

(x + y) 1 x+y
= × =
(x − y) 1 x−y

Final answer:

x+y
x−y

(b) Solve

(i) x2 + x − 6 > 0

Solution:
Factor
x2 + x − 6 = (x + 3)(x − 2)

Set the inequality:

(x + 3)(x − 2) > 0
Sign analysis:

Page 11 of 14
 • Critical points: x = −3 and x = 2
• Test intervals:
i. x < −3: pick x = −4:

(−4 + 3)(−4 − 2) = (−1)(−6) = 6 > 0

ii. −3 < x < 2: pick x = 0:

(0 + 3)(0 − 2) = (3)(−2) = −6 < 0

iii. x > 2: pick x = 3:

(3 + 3)(3 − 2) = (6)(1) = 6 > 0

Final answer:

x ∈ (−∞, −3) ∪ (2, ∞)

x+1
(ii) x−1 >0

Solution:

Solve the inequality:

x+1
x−1 > 0
Critical points: x = −1 (numerator zero), x = 1 (denominator zero).
Number line analysis:
• For x < −1, pick x = −2:
−2 + 1 −1 1
= = >0
−2 − 1 −3 3
• For −1 < x < 1, pick x = 0:
0+1 1
= = −1 < 0
0−1 −1
• For x > 1, pick x = 2:
2+1 3
= =3>0
2−1 1
Final answer:

x ∈ (−∞, −1) ∪ (1, ∞)

[8,4,3]

Page 12 of 14
QUESTION 4

(a) Define the following terms

(i) Quadratic equation

Solution.

A quadratic equation is a second-degree polynomial equation in the form:

ax2 + bx + c = 0,

where a 6= 0, and b, c are real coefficients.

(ii) Polynomial function

Solution.

A polynomial function is a function that can be expressed as:

f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 ,

where an 6= 0, and n is a non-negative integer.

(iii) Modulus function

Solution

The modulus function |x| is defined as:

(
x, if x ≥ 0,
|x| =
−x, if x < 0.

It measures the distance of x from zero on the real number line.

1 10
(b) Show that if a + a = 3 then 3a2 − 10a + 3 = 0. Hence,find the possible values of a.

Solution

Step 1: Given:
1 10
a+ =
a 3
Multiply both sides by a:
10
a2 + 1 = a
3
Rearrange:
10
a2 − a+1=0
3

Page 13 of 14
 Multiply the entire equation by 3 to clear denominators:

3a2 − 10a + 3 = 0

This is the required quadratic in a.

Final answer:

3a2 − 10a + 3 = 0

Possible values of a :
Use quadratic formula:
p
10 ± (−10)2 − 4 × 3 × 3
a=
2×3
√
10 ± 100 − 36
=
√6
10 ± 64
=
6
10 ± 8
=
6
Therefore, the two solutions are:

10 + 8 18
a= = =3
6 6
10 − 8 2 1
a= = =
6 6 3
Final answer:

1
a = 3, a =
3
[9,6]

♠End of Mathematics marking key for CAT 2♠

”please make use of these solutions”

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