Quadratic Equations, SPM Practice (Paper 2)

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Question 4:
It is given α and β are the roots of the quadratic equation x (x – 3) = 2k – 4, where k is a constant.
 (a) Find the range of values of k if α≠β. (b) Given  α 2  and  β 2  are the roots of another quadratic equation      2 x 2 +tx−4=0, where t is a constant, find the value of t and of k.

Solution:
(a) x( x−3 )=2k−4 x 2 −3x+4−2k=0 a=1, b=−3, c=4−2k                     b 2 −4ac>0 ( −3 ) 2 −4( 1 )( 4−2k )>0                9−16+8k>0                             8k>7                               k> 7 8

(b) From the equation  x 2 −3x+4−2k=0, α+β=− b a          =− −3 1          =3.............( 1 ) αβ= c a     = 4−2k 1     =4−2k.............( 2 ) From the equation 2 x 2 +tx−4=0, α 2 + β 2 =− t 2 α+β=−t.............( 3 ) α 2 × β 2 =− 4 2 αβ=−8.............( 4 ) Substitute (1)=(3), 3=−t t=−3 Substitute (2)=(4), 4−2k=−8 4+8=2k k=6

Quadratic Equations, SPM Practice (Paper 2)

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2.10.2 Quadratic Equations, SPM Practice (Paper 2)

Question 3:
If α and β are the roots of the quadratic equation 3x2 + 2x– 5 = 0, form the quadratic equations that have the following roots.
(a)  2 α  and  2 β (b)  ( α + 2 β )  and  ( β + 2 α )

Solution:
3x2 + 2x – 5 = 0
a = 3, b = 2, c = –5
The roots are α and β.
α + β = − b a = − 2 3 α β = c a = − 5 3

(a)
The new roots are  2 α and 2 β . Sum of new roots = 2 α + 2 β = 2 β + 2 α α β = 2 ( α + β ) α β = 2 ( − 2 3 ) − 5 3 = 4 5

Product of new roots = ( 2 α ) ( 2 β ) = 4 α β = 4 − 5 3 = − 12 5

Using the formula, x2– (sum of roots)x + product of roots = 0
The new quadratic equation is
x 2 − ( 4 5 ) x + ( − 12 5 ) = 0
5x2 – 4x– 12 = 0



(b)
The new roots are  ( α + 2 β ) and ( β + 2 α ) . Sum of new roots = ( α + 2 β ) + ( β + 2 α )
= α + β + ( 2 α + 2 β ) = α + β + 2 α + 2 β α β = α + β + 2 ( α + β ) α β = 4 5 + 2 ( 4 5 ) − 12 5 = 4 5 − 2 3 = 2 15
Product of new roots = ( α + 2 β ) ( β + 2 α ) = α β + 2 + 2 + 4 α β
= − 12 5 + 4 + 4 − 12 5 = − 12 5 + 4 − 5 3 = − 1 15

The new quadratic equation is
x 2 − ( 2 15 ) x + ( − 1 15 ) = 0
15x2 – 2x– 1 = 0

Quadratic Equations, SPM Practice (Paper 2)

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Question 2:
Given α and β are two roots of the quadratic equation (2x + 5)(x + 1) + p = 0 where αβ = 3 and p is a constant.
Find the value p, α and of β.

Solutions:
(2x + 5)(x + 1) + p = 0
2x2 + 2x + 5x + 5 + p = 0
2x2 + 7x + 5 + p = 0
*Compare with, x2– (sum of roots)x + product of roots = 0
x 2 + 7 2 x + 5 + p 2 = 0 ← divide both  sides with 2
Product of roots, αβ = 3
5 + p 2 = 3  
5 + p = 6
p = 1

Sum of roots = − 7 2  
   α + β = − 7 2   →  (1) and  α β = 3    →  (2) from (2),  β = 3 α    →  (3) Substitute (3) into (1), α + 3 α = − 7 2  

2α2+ 6 = –7α ← (multiply both sides with 2α)
2α2+ 7α + 6 = 0
(2α + 3)(α + 2) = 0
2α + 3 = 0   or α + 2 = 0
α=− 3 2    α = –2

Substitute  α = − 3 2  into (3), β = 3 − 3 2 = 3 ( − 2 3 ) = − 2

Substitute α = –2 into (3),
β = − 3 2 ∴   p = 1 ,  and when  α = − 3 2 , β = − 2  and  α = − 2 , β = − 3 2 .


Quadratic Equations, SPM Practice (Paper 2)

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2.10.1 Quadratic Equations, SPM Practice (Paper 2)

Question 1:
(a)  Find the values of k if the equation (1 – k) x2– 2(k + 5)x + k + 4 = 0 has real and equal roots.
Hence, find the roots of the equation based on the values of k obtained.
(b)  Given the curve y = 5 + 4x – x2 has tangen equation in the form y = px + 9. Calculate the possible values of p.

Solutions:
(a)
For equal roots,
b2 – 4ac = 0
[–2(k + 5)] 2 – 4(1 – k)( k + 4) = 0
4(k + 5) 2 – 4(1 – k)( k + 4) = 0
4(k2 + 10k + 25) – 4(4 – 3k – k2) = 0
4k2 + 40k + 100 – 16 + 12k + 4k2 = 0
8k2 + 52k + 84 = 0
2k2 + 13k + 21 = 0
(2k + 7) (k + 3) = 0
k = − 7 2 ,   − 3

If k = − 7 2  , the equation is
( 1 + 7 2 ) x 2 − 2 ( − 7 2 + 5 ) x − 7 2 + 4 = 0 9 2 x 2 − 3 x + 1 2 = 0  

9x2 – 6x + 1 = 0
(3x – 1) (3x – 1) = 0
x = ⅓

If k = –3, the equation is
(1 + 3)x 2 – 2(–3 + 5)x – 3 + 4 = 0
4x2 – 4x + 1 = 0
(2x – 1) (2x – 1) = 0
x = ½

(b)
y = 5 + 4x – x2 ----- (1)
y = px + 9 ---------- (2)
(1)  = (2), 5 + 4x – x2= px + 9
x2 + px – 4x + 9 – 5 = 0
x2 + (p – 4)x + 4 = 0

Tangen equation only has one intersection point with equal roots.
b2 – 4ac = 0
(p – 4)2 – 4(1)(4) = 0
p2 – 8p + 16 – 16 = 0
p2 – 8p = 0
p (p – 8) = 0
Therefore, p = 0 and p = 8.

2.2a Solving Quadratic Equations – Factorisation

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2.4.1 Solving Quadratic Equations – Factorisation
1. If a quadratic equation can be factorised into a product of two factors such that

(x – p)(x – q) = 0

Hence
 x – p = 0   or  x – q = 0
   x = p   or x = q

p and q  are the roots of the equation.

Notes
1.The equation must be written in general form ax2 + bx+ c = 0 before factorisation.
2. This method can only be used if the quadratic expression can be factorised completely.



Example 1:
Find the roots of the quadratic equations
(a) x (2x − 8) = 0 
(b) x2 −16x = 0
(c) 3x2 − 75x = 0
(d) 5x2 − 100x = 25x

Solution:
(a) 
x (2x − 8) = 0 
x = 0  or  2x − 8 = 0
2x − 8 = 0
2x = 8
x= 4
x = 0  or  x = 4

(b)
x2 −16x = 0
x (x − 16) = 0 
x = 0  or  x − 16 = 0
x = 0  or  x = 16

(c) 
3x2 − 75x = 0
3x (x − 25) = 0 
3x = 0  or  x − 25 = 0
x = 0  or  x = 25

(d) 
5x2 − 100x = 25x
5x2 − 100x − 25x = 0
5x2 − 125x = 0
x (5x − 125) = 0 
x = 0  or  5x − 125 = 0
5x = 125
x = 25
x = 0  or  x = 25



Example 2:
Solve the following quadratic equations
(a) x2 − 4x – 5 = 0
(b) 1 − 5x + 2x2 = 4

Solution:
(a) 
x2 − 4x – 5 = 0
(x – 5) (x + 1) = 0
x – 5 = 0  or  x + 1 = 0
x = 5  or  x = –1

(b)
1 − 5x + 2x2 = 4
2x2 − 5x + 1 – 4 = 0
2x2 − 5x – 3 = 0
(2x + 1) (x – 3) = 0
2x + 1= 0  or  x – 3 = 0
2x = –1  or  x = 3
x = –½  or  x = 3

Roots of Quadratic Equations

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Roots of Quadratic Equations

Roots of a quadratic equation are the values of variables/unknowns that satisfy the equation.

Example:
Determine whether 1, 2, and 3 are the roots of the quadratic equation x 2 − 5 x + 6 = 0 .

Answer:
When x = 1,
x 2 − 5 x + 6 = 0 ( 1 ) 2 − 5 ( 1 ) + 6 = 0 2 = 0
x = 1 does not satisfy the equation

When x = 2,
x 2 − 5 x + 6 = 0 ( 2 ) 2 − 5 ( 2 ) + 6 = 0 0 = 0
x = 2 satisfies the equation.

When x = 3
x 2 − 5 x + 6 = 0 ( 3 ) 2 − 5 ( 3 ) + 6 = 0 0 = 0
x = 3 satisfies the equation.

 Conclusion:
  1. 2 and 3 satisfy the equation x 2 − 5 x + 6 = 0 , hence there are the roots of the equation.
  2. 1 does not satisfy the equation x 2 − 5 x + 6 = 0 , hence it is NOT the root of the equation.

SPM Practice (Long Question)

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Question 3:
Given that f : x → hx + k and f2 : x → 4x + 15.  
(a) Find the value of h and of k.
(b) Take the value of h > 0, find the values of x for which f (x2 ) = 7x

Solution:
(a)
Given f (x) = hx + k
f2 (x) = ff (x) = f (hx + k)
= h (hx + k) + k
= h2 x + hk + k

f2 (x) = 4x + 15
h2 x + hk + k = 4x + 15
h2 = 4
h = ± 2
when, h = 2
hk + k = 15
2k + k = 15
k = 5

When, h = –2
hk + k = 15
–2k + k = 15
k = –15

(b)
h > 0, h = 2, k = 5
Given f (x) = hx + k
f (x) = 2x + 5 f (x2 ) = 7x
2 (x2 ) + 5 = 7x
2x2 – 7x + 5 = 0
(2x – 5)(x –1) = 0
2x – 5 = 0 or x –1= 0
x = 5/2 x = 1

SPM Practice (Long Question)

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Question 1:
The function f and g is defined by
f:x↦2x−3 g:x↦ 2 x ;x≠0
Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f-1 ,
Calculate the value of x such that ff(x) = gf(x).

Solution:
(a)
ff( x )=f[ f( x ) ]          =f( 2x−3 )          =2( 2x−3 )−3          =4x−9 ff:x↦4x−9

(b)
gf( x )=g[ f( x ) ]           =g( 2x−3 )           = 2 2x−3 gf:x↦ 2 2x−3

(c)
Let  f −1 ( x )=y, thus  f( y )=x       2y−3=x               y= x+3 2 ∴   f −1 ( x )= x+3 2 f −1 :x↦ x+3 2 When ff( x )=gf( x ), 4x−9= 2 2x−3 ( 4x−9 )( 2x−3 )=2 8 x 2 −30x+27=2 8 x 2 −30x+25=0 ( 4x−5 )( 2x−5 )=0 4x−5=0       or       2x−5=0 x= 5 4              or              x= 5 2

SPM Practice (Short Question)

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Question 1:
Given the function g : x → 3x – 2, find  
(a) the value of x when g(x) maps onto itself,
(b) the value of k such that g(2 – k) = 4k.

Solution:
(a)
  g( x )=x 3x−2=x 3x−x=2  2x=2 x=1

(b)
  g( x )=3x−2 g( 2−k )=4k 3( 2−k )−2=4k 6−3k−2=4k    −7k=−4  k= 4 7



Question 2:
Given the functions f : x → px + 1, g : x → 3x – 5 and fg(x) = 3px + q.  
Express p in terms of q.

Solution:
f( x )=px+1, g( x )=3x−5 fg( x )=p( 3x−5 )+1  =3px−5p+1 Given fg( x )=3px+q 3px−5p+1=3px+q   −5p+1=q    −5p=q−1   5p=1−q p= 1−q 5



Question 3:
Given the functions h : x → 3x + 1, and gh : x → 9x2 + 6x – 4, find  
(a) h-1 (x),
(b) g(x).

Solution:
(a)
Let  h −1 ( x )=y, thus  h( y )=x    3y+1=x 3y=x−1   y= x−1 3 ∴   h −1 ( x )= x−1 3 h −1 :x↦ x−1 3

(b)
g[ h( x ) ]=9 x 2 +6x−4 g( 3x+1 )=9 x 2 +6x−4 Let y=3x+1 thus  x= y−1 3  g( y )=9 ( y−1 3 ) 2 +6( y−1 3 )−4 = 9 ( y−1 ) 2 9 +2( y−1 )−4 = y 2 −2y+1+2y−2−4 = y 2 −5 ∴ g( x )= x 2 −5

Inverse Function Example 4

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Example 4:
(a) If f:x→x−2, find  f −1 ( 5 ),
(b) if  f : x → x + 9 x − 5 ,   x ≠ 5 ,  find  f − 1 ( 3 ) .

Solution:
(a)
f (x) = x– 2
Let y = f -1 (5)
f (y) = 5
y – 2 = 5
y = 7
therefore, f -1 (5) = 7

(b)
f ( x ) = x + 9 x − 5 Let  y = f − 1 ( 3 ) f ( y ) = 3 y + 9 y − 5 = 3 y + 9 = 3 y − 15 2 y = 24 y = 12 ∴ f − 1 ( 3 ) = 7