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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


Composite Function (Comparison Method) Example 3

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

Solution:
(a)
Step 1:
Find f2 (x)
Given f (x) = hx + k
f2 (x) = ff (x) = f (hx + k)
= h (hx + k) + k
= h2x + hk + k

Step 2:
Compare with given f2 (x)
f2 (x) = 4x + 15
h2x + 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
 or
x = 1

Example 1

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Example 1
Given the function f : x → 6 x + 1 . Find the value of p if f ( 4 ) = 4 p + 5 .

Answer:
f : x → 6 x + 1 f ( x ) = 6 x + 1 f ( 4 ) = 6 ( 4 ) + 1 f ( 4 ) = 25

f ( 4 ) = 4 p +   5 25 = 4 p +   5 4 p = 25 − 5 = 20 p = 20 4 = 5

4.1 Simultaneous Equations

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4.1 Simultaneous Equations

(A) Steps in solving simultaneous equations:
  1. For the linear equation, arrange so that one of the unknown becomes the subject of the equation.
  2. Substitute the linear equation into the non-linear equation.
  3. Simplify and expressed the equation in the general form of quadratic equation a x 2 + b x + c = 0
  4. Solve the quadratic equation. 
  5. Find the value of the second unknown by substituting the value obtained into the linear equation.

Example:
Solve the following simultaneous equations.
y + x = 9 x y = 20

Solution:
For the linear equation, arrange so that one of the unknown becomes the subject of the equation.
y + x = 9 y = 9 − x

Substitute the linear equation into the non-linear equation.
x y = 20 x ( 9 − x ) = 20 9 x − x 2 = 20

Simplify and expressed the equation in the general form of quadratic equation a x 2 + b x + c = 0
9 x − x 2 = 20 x 2 − 9 x + 20 = 0

Solve the quadratic equation. 
x 2 − 9 x + 20 = 0 ( x − 4 ) ( x − 5 ) = 0 x = 4  or  x = 5

Find the value of the second unknown by substituting the value obtained into the linear equation.
When  x = 4 , y = 9 − x = 9 − 4 = 5 When  x = 5 , y = 9 − x = 9 − 5 = 4