Page 8 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Chapter 2 Quadratic Functions

Solution Solution
(a) Rearrange 3x = 2x + 5 to the general form of Applying the Pythagoras theorem,
2
ax + bx + c = 0. AC = AB + BC 2
2
2
2
3x – 2x – 5 = 0 (p + 4) = (p – 3) + (p + 1) 2
2
2
2
2
2
2
Therefore, a = 3, b = –2 and c = –5. p + 8p + 16 = p – 6p + 9 + p + 2p + 1
2
Substitute the values of a, b and c into the quadratic p – 12p – 6 = 0 –(–12) ± (–12) – 4(1)(–6)
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2
formula, p = 2(1)
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2

x = –(–2) ± (–2) – 4(3)(–5) 12 ± 168
2(3) =
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
2 ± 4 + 60 –b ± b – 4ac 2
2
= 6 x = 2a = 12 + 168 or 12 – 168
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
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2 ± 64 2 2
= Form 4
6 = 12.48 or –0.4807 (not acceptable)
2 ± 8
=
6 Therefore, p = 12.48
x = 2 + 8 or x = 2 – 8
6 6 Try Questions 3 – 7 in ‘Try This! 2.1’
5
= or = –1
3
C
C C A L C U L A T O R Corner
(b) Rearrange 6 – 2x = 5x to the general form of Solve the quadratic equation x – 14x – 39 = 0.
2
2
ax + bx + c = 0.
2
5x + 2x – 6 = 0 1: Press the MODE key until EQN is displayed.
2
Therefore, a = 5, b = 2 and c = –6. 2: Press 1
Substitute the values of a, b and c into the quadratic 3: ‘Unknowns?’ is displayed, press 
formula, 4: ‘Degree?’ is displayed, press 2
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2
–2 ± (2) – 4(5)(–6) 5: ‘a?’ is displayed, press 1 =
x =
2(5) 6: ‘b?’ is displayed, press – 1 4 =
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
–2 ± 124 –b ± b – 4ac
2
= x = 7: ‘c?’ is displayed, press – 3 9 =
10 2a 8: One of the roots is, ‘x = 16.38’ is displayed
1


–2 + 124 –2 – 124 8: Press the  key. Another root is, ‘x = –2.381’ is
x = or x = 2
10 10 displayed
= 0.9136 or = –1.314
Try Question 2 in ‘Try This! 2.1’ B Forming quadratic equations from
given roots
1. If α and β are the roots of a quadratic equation,
3 then
A
x = α and x = b
(x – α) = 0 and (x – b) = 0
(p + 4) cm
(p – 3) cm (x – α)(x – b) = 0
x – αx – bx + αb = 0
2
x – (α + b)x + αb = 0
2
C
B (p + 1) cm
α + b is called the sum of the roots (SOR) and αb
The diagram above shows a right-angled triangle is called the product of the roots (POR).
ABC with AB = (p – 3) cm, BC = (p + 1) cm and
AC = (p + 4) cm. Find the value of p. [Give your answer x – (SOR)x + (POR) = 0
2
correct to 4 significant figures.]
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