Chapter 1 Quadratic Equations & Quadratic Functions
CHAPTER 1
5 89 5 89 − )( x − + )=0 4 4 4 4 5 + 89 5 − 89 or x= 4 4
(x −
bñÉêÅáëÉ=N^=EéKUF 1. (a) x 2 − 2 x + 1 = 0 ( x − 1)2 = 0 x =1
x − 4x − 5 = 0 ( x − 5)( x + 1) = 0 x = 5 or −1 2 (c) 3 x − 7 x + 4 = 0 (3 x − 4)( x − 1) = 0 4 x = or 1 3
(b)
(d)
2
8x 2 = 6 x − 1 8x 2 − 6 x + 1 = 0 ( 4 x − 1)(2 x − 1) = 0 1 1 x = or 4 2
(e) 56 − 10 x − 6 x = 0 2
6 x 2 + 10 x − 56 = 0 3 x 2 + 5 x − 28 = 0 (3 x − 7)( x + 4) = 0 7 or −4 x= 3
(f)
2(5 x 2 − 1) − x = 0 10 x 2 − 2 − x = 0 10 x 2 − x − 2 = 0 (5 x + 2)(2 x − 1) = 0 x=−
2. (a)
2 1 or 5 2
x2 + 6x + 5 = 0 6 6 x 2 + 6 x + ( )2 − ( )2 + 5 = 0 2 2 ( x + 3)2 − 9 + 5 = 0 ( x + 3)2 = 4 x + 3 = ±2 x = −5 or −1
(b)
2 x 2 − 5x − 8 = 0 5 x2 − x − 4 = 0 2 5 2 5 2 2 5 x − x +( ) −( ) −4 = 0 2 4 4 5 2 25 (x − ) − −4=0 4 16 5 89 ( x − )2 − =0 4 16 5 89 2 ( x − )2 − ( ) =0 4 4
3. (a) 2 x 2 − 5 x − 6 = 0 Let a = 2, b = −5, c = −6. Using the formula,
−( −5) ± ( −5)2 − 4(2)( −6) 2( 2 ) 5 ± 25 + 48 = 4 5 ± 73 = 4
x=
(b) 35 − 100 x + x 2 = 0
x 2 − 100 x + 35 = 0 Let a = 1, b = −100, c = 35. Using the formula, −( −100) ± ( −100)2 − 4(1)(35) 2(1) 100 ± 10 000 − 140 = 2 100 ± 9 860 = 2 = 50 ± 2 465
x=
(c) 2 x 2 + 4 x = 5 + 5 x
2x2 − x − 5 = 0 Let a = 2, b = −1, c = −5. Using the formula, −( −1) ± ( −1)2 − 4(2)( −5) 2( 2 ) 1 ± 1 + 40 = 2( 2 ) 1 ± 41 = 4
x=
(d) 2 x 2 + 4 x = −3
2x2 + 4x + 3 = 0 Let a = 2, b = 4, c = 3. Using the formula, −4 ± 4 2 − 4(2)(3) −4 ± −8 = 2( 2 ) 4 ∴ The equation has no real solution. x=
17
18
Chapter 1 Quadratic Equations & Quadratic Functions
x2 x −1 = 3 6 2x2 − 6 = x
(e)
x −2 − x −1 − 56 = 0
8.
( x −1 − 8)( x −1 + 7) = 0
x −1 = 8 or −7 1 1 x= or − 8 7
2x − x − 6 = 0 2
Let a = 2, b = −1, c = −6. Using the formula, −( −1) ± ( −1) − 4(2)( −6) 2( 2 ) 1 ± 49 = 4 1± 7 = 4 −3 or 2 = 2 2
x=
1
2x − 9x 2 + 4 = 0
9.
1
1
(2 x 2 − 1)( x 2 − 4) = 0 1 1 x2 = or 4 2 1 or 16 x= 4 2y = 5 +
10.
(f) 3 x 2 + 5 x − 1 = x ( x + 2)
2 y 2 = 5y + 3
3x 2 + 5x − 1 = x 2 + 2 x
2 y − 5y − 3 = 0 (2 y + 1)( y − 3) = 0 2
2 x 2 + 3x − 1 = 0
Let a = 2, b = 3, c = −1. Using the formula,
y=−
−3 ± 32 − 4(2)( −1) 2( 2 ) −3 ± 17 = 4
x=
4.
11.
(3 x + 7)( x − 1) = 4( x − 1) (3 x + 7)( x − 1) − 4( x − 1) = 0 ( x − 1)(3 x + 7 − 4) = 0 ( x − 1)(3 x + 3) = 0 x = 1 or −1
2
1 or −2 5
x 2 − 2 x − 35 = 0 ( x − 7)( x + 5) = 0 x = 7 or −5 8p 2 8p 2 ) − ( ) +1 2 2 = ( x + 4 p)2 − 16 p 2 + 1
12. (a) x 2 + 8 px + 1 = x 2 + 8 px + (
1
or 4 x3 = 9 x = 729 or 64
5 is a root of 2 x 2 − 3 x − a = 0 , 2 5 5 2( )2 − 3( ) − a = 0 2 2 a=5
13. Since
1
( x 3 − 9)( x 3 − 4) = 0
l = 4 p , m = 1 − 16 p 2
p p (b) x 2 − px + 2 p = x 2 − px + ( )2 − ( )2 + 2 p 2 2 p 2 p2 = (x − ) − + 2p 2 4 p p2 = ( x − ) 2 + (2 p − ) 2 4 p p2 ∴ l = − , m = 2p − 2 4
1
x 3 − 13 x 3 + 36 = 0 1
3 x 2 − 11x − 40 = x−5 2x + 1 3 x 2 − 11x − 40 = 2 x 2 − 9 x − 5
∴
6. ( x 2 + 5 x )2 − 8( x 2 + 5 x ) − 84 = 0 ( x 2 + 5 x + 6)( x 2 + 5 x − 14) = 0 ( x + 2)( x + 3)( x − 2)( x + 7) = 9 x = −3, −2, 2, −7 7.
1 or 3 2
= ( x + 4 p)2 + (1 − 16 p 2 )
5. 5(2 x + 1)2 + 12(2 x + 1) − 9 = 0 [5(2 x + 1) − 3][(2 x + 1) + 3] = 0 (10 x + 5 − 3)(2 x + 1 + 3) = 0 (10 x + 2)(2 x + 4) = 0 x=−
3 y
∴
Chapter 1 Quadratic Equations & Quadratic Functions
14. Since −2 is a root of 2 x 2 + 6 x + k = 0 , 2( −2) + 6( −2) + k = 0 k=4 2
Substitute k = 4 into the given equation, 2x2 + 6x + 4 = 0 x 2 + 3x + 2 = 0 ( x + 1)( x + 2) = 0 ∴ x = −1 or − 2 ∴ The other root of the equation is −1 . 5 3 x+ ) 2 2 5 5 5 3 = 2[ x 2 − x + ( )2 − ( )2 + ] 2 4 4 2 1 5 = 2( x − ) 2 + 2( − ) 4 16 5 1 = 2( x − ) 2 − 4 8
15. 2 x 2 − 5 x + 3 = 2( x 2 −
3 is a root of (*), 4 ( − 43 )2 + 3( − 43 ) + 1 =l 2( − 43 )2 − 5( − 43 ) + 2 1 l=− 10 1 (c) Substitute l = − into the equation obtain 10 in (a), 1 1 [2( − ) − 1]x 2 − [5( − ) + 3]x 10 10 1 +[2( − ) − 1] = 0 10 12 25 12 − x2 − =0 x− 10 10 10 (b) Since −
12 x 2 + 25 x + 12 = 0 (3 x + 4)( 4 x + 3) = 0 x=−
4 3
or
−
3 4
4 ∴ The other root of equation is − . 3
16. (a) Since 2 is a root of
m 2 x 2 + 2(2 m − 5) x + 8 = 0 . m 2 (2)2 + 2(2 m − 5)(2) + 8 = 0
bñÉêÅáëÉ=N_=EéKNPF
4 m + 8m − 20 + 8 = 0 2
1. kx 2 − (3k − 5) x + 14 = 0
4 m + 8m − 12 = 0 2
m2 + 2m − 3 = 0 ( m − 1)( m + 3) = 0 m = 1 or −3
Let a = k , b = −(3k − 5) , c = 14 D = [ −(3k − 5)]2 − 4( k )(14) = 9k 2 − 30 k + 25 − 56k
(b) When m = 1,
= 9k 2 − 86k + 25
x 2 + 2(2 − 5) x + 8 = 0
2. (2 k + 1) x 2 − 2 k = ( k + 4) x
x2 − 6x + 8 = 0 ( x − 2)( x − 4) = 0 x = 2 or 4
(2 k + 1) x 2 − ( k + 4) x − 2 k = 0 Let a = 2 k + 1 , b = −( k + 4) , c = −2 k
∴ When m = 1, the other root = 4 When m = −3 ,
D = [ −( k + 4)]2 − 4(2 k + 1)( −2 k ) = k 2 + 8k + 16 + 16k 2 + 8k
9 x 2 + 2( −6 − 5) x + 8 = 0
= 17k 2 + 16k + 16
9 x 2 − 22 x + 8 = 0 (9 x − 4)( x − 2) = 0 x = 2 or
3. x 2 − kx + 16 = 0
4 9
4 ∴ When m = −3 , the other root = 9
D = k 2 − 4(16) = 0 k 2 = 64 k = ±8
4. x 2 − k ( x − 1) = 0 x + 3x + 1 =l 2 x 2 − 5x + 2 x 2 + 3 x + 1 = 2lx 2 − 5lx + 2l 2
17. (a)
(2l − 1) x 2 − (5l + 3) x + (2l − 1) = 0
19
x 2 − kx + k = 0 ∴
D = k 2 − 4k = 0 k ( k − 4) = 0 k = 0 or 4
20
Chapter 1 Quadratic Equations & Quadratic Functions
If m ≠ n , D < 0 . ∴ The equation has unreal roots.
5. kx 2 − 2 x + 1 = 0
D = ( −2)2 − 4 k = 0
4 − 4k = 0 4k = 4 k =1
11. x 2 + 2 ax + a 2 − b 2 − c 2 = 0 D = ( 2 a ) 2 − 4( a 2 − b 2 − c 2 )
6. x 2 − kx + k + 3 = 0 D = ( − k )2 − 4( k + 3) = 0
= 4 a 2 − 4 a 2 + 4 b 2 + 4c 2 = 4( b 2 + c 2 )
k − 4 k − 12 = 0 ( k − 6)( k + 2) = 0 k = 6 or −2 2
If b ≠ 0 and c ≠ 0 , D > 0 . ∴ The equation has unequal real roots.
7. x 2 − 2 ax + a 2 − b 2 − 2 bc − c 2 = 0 D = ( −2 a)2 − 4( a 2 − b 2 − 2 bc − c 2 ) = 4 a 2 − 4 a 2 + 4b 2 + 8bc + 4c 2
12. ( x − a)( x − b) = c x 2 − ( a + b) x + ab − c = 0
D = ( a + b)2 − 4( ab − c)
= 4b 2 + 8bc + 4c 2
= a 2 + 2 ab + b 2 − 4 ab + 4c
= 4(b 2 + 2 bc + c 2 )
= a 2 − 2 ab + b 2 + 4c
= 4( b + c ) 2 = [2(b + c)]2 which is a perfect square for any integers b and c. ∴ Roots are rational.
8. px 2 + 2 qx − p + 2 q = 0
= ( a − b ) 2 + 4c Q
( a − b)2 ≥ 0 and c > 0
∴ ( a − b ) 2 + 4c > 0 ∴ The roots of the equation are real.
D = (2 q )2 − 4 p( − p + 2 q ) = 4 q 2 + 4 p 2 − 8 pq
13. x 2 − 2( a + 3) x + (11a + 3) = 0
= 4( p 2 − 2 pq + q 2 )
D = [ −2( a + 3)]2 − 4(11a + 3) =0
= 4( p − q )
2
= [2( p − q )]2 which is the square of a rational number as p and q are rational. ∴ Roots are rational.
9. x 2 − ( a + b + c) x + a(b + c) = 0 D = ( a + b + c ) 2 − 4 a( b + c ) 2
10. ( m + n ) x − 2( m + n) x + 2 = 0 2
D = [ −2( m + n)] − 4( m + n )(2) 2
2
2
= 4( m + 2 mn + n ) − 8m − 8n 2
2
= 4 m 2 + 8mn + 4n 2 − 8m 2 − 8n 2 = −4 m 2 + 8mn − 4n 2 = −4( m 2 − 2 mn + n 2 ) = −4( m − n)2
∴ When a = 2 , x = 5
x 2 − 12 x + 36 = 0
= 4( m + n)2 − 8( m 2 + n 2 ) 2
( x − 5)2 = 0 x=5
When a = 3 , the equation is
2
2
a 2 − 5a + 6 = 0 ( a − 2)( a − 3) = 0 a = 2 or 3 x 2 − 10 x + 25 = 0
= 4a − 4a =0 ∴ The equation has equal roots. 2
4( a 2 − 5a + 6 ) = 0
When a = 2 , the equation is
= ( a + a)2 − 4 a ⋅ a 2
4( a 2 + 6 a + 9) − 4(11a + 3) = 0
( x − 6) 2 = 0 x=6
∴ When a = 3 , x = 6
Chapter 1 Quadratic Equations & Quadratic Functions
D = 16 − 4(c − 1)(3 − c)
14. x 2 + 2( a + 2) x + (5a + 24) = 0
= 16 − 4(3c − c 2 − 3 + c)
D = [2( a + 2)]2 − 4(5a + 24) = 0
= 4[ 4 − ( − c 2 + 4c − 3)]
4( a 2 + 4 a + 4) − 4(5a + 24) = 0
= 4( 4 + c 2 − 4c + 3)
4( a − a − 20) = 0 2
Q 4(c 2 − 4c + 7)
a 2 − a − 20 = 0 ( a − 5)( a + 4) = 0 a = 5 or
= 4[(c 2 − 4c + 4) + 3]
−4
When a = 5 , the equation is
= 4[(c − 2)2 + 3] > 0 Hence, if (*) has unequal real roots, c can be any real numbers. But for (*) to be a quadratic equation, a ≠ c . ∴ c can be any real number other than 1.
x 2 + 14 x + 49 = 0 ( x + 7)2 = 0 x = −7 ∴ When a = 5 , x = −7
bñÉêÅáëÉ=N`=EéKNVF
When a = −4 , the equation is
1. (a) x 2 − 4 x + 1 = 0
x2 − 4x + 4 = 0
−4 =4 1 1 Product of the roots = = 1 1 2 (b) 48 x = 22 x + 15 Sum of the roots = −
( x − 2)2 = 0 x=2 ∴ When a = −4 , x = 2 15. x 2 + ( m + n) x + ( m 2 − n 2 ) = 0
48 x 2 − 22 x − 15 = 0
D = ( m + n ) − 4( m − n ) = 0
Sum of the roots =
2
2
22 11 = 48 24 15 5 Product of the roots = − =− 48 16
2
m 2 + 2 mn + n 2 − 4( m 2 − n 2 ) = 0 −3m 2 + 2 mn + 5n 2 = 0 ( −3m + 5n)( m + n) = 0 5n m= 3
or − n
16. (a) (c − a) x 2 − 2( a − b) x + (b − c) = 0 D = [ −2( a − b)] − 4(c − a)(b − c) 2
= 4( a − b)2 − 4(c − a)(b − c) = 4[( a − b)2 − (c − a)(b − c)] = 4[a 2 − 2 ab + b 2 − (bc − c 2 − ab + ac)] = 4( a 2 + b 2 + c 2 − ab − bc − ac) = 2(2 a 2 + 2 b 2 + 2c 2 − 2 ab − 2 bc − 2ca) = 2( a 2 − 2 ab + b 2 + b 2 − 2 bc + c 2 + c 2 − 2ca + a 2 ) = 2[( a − b)2 + (b − c)2 + (c − a)2 ] Q a, b and c are not all equal.
∴ 2[( a − b)2 + (b − c)2 + (c − a)2 ] > 0 ∴ The equation has unequal real roots. (b) If a = 1 , b = 3 , the equation becomes
(c − 1) x 2 − 2( −2) x + (3 − c) = 0 (c − 1) x 2 + 4 x + (3 − c) = 0
21
(c) ( 4 x + 7)2 = 4 x 2 16 x 2 + 56 x + 49 − 4 x 2 = 0 12 x 2 + 56 x + 49 = 0
56 14 =− 12 3 49 Product of the roots = 12 Sum of the roots = −
(d) 6 − 5 x − 25 x 2 = 0 25 x 2 + 5 x − 6 = 0 5 1 =− Sum of the roots = − 25 5 6 Product of the roots = − 25
2. 3 x 2 − 5 x − 1 = 0 5 1 α + β = and αβ = − 3 3 2 2 2 (a) α + β = (α + β) − 2αβ 25 2 = + 9 3 31 = 9
22
Chapter 1 Quadratic Equations & Quadratic Functions
(b)
(d) α 4 + β 4 = α 4 + 2α 2β 2 + β 4 − 2α 2β 2
α β α 2 + β2 + = β α αβ 31 = 9 1 − 3 31 =− 3
= (α 2 + β 2 )2 − 2(αβ)2 = [(α + β)2 − 2αβ]2 − 2(αβ)2 = [( − p)2 − 2 q]2 − 2 q 2 = ( p 2 − 2 q)2 − 2 q 2 = p 4 − 4 p2 q + 4q 2 − 2q 2 = p 4 − 4 p2 q + 2q 2
(c) (1 − α )(1 − β) = 1 − (α + β) + αβ 5 1 = 1− − 3 3 = −1
3 2 5 + =− 2 3 6 3 2 Product of the roots = ( − )( ) = −1 2 3 5 One required equation is x 2 + x − 1 = 0 , 6 i.e. 6 x 2 + 5 x − 6 = 0 .
(b) Sum of the roots = −
3. x 2 + px + q = 0 α + β = − p and αβ = q
(a) α − αβ + β = α + 2αβ + β − 3αβ 2
2
Product of the roots = 4( −5) = −20 One required equation is x 2 + x − 20 = 0 .
(d) αβ 2 + α 2β = αβ(α + β) 1 5 =− ( ) 3 3 5 =− 9
2
4. (a) Sum of the roots = 4 + ( −5) = −1
2
= (α + β)2 − 3αβ
(c) Sum of the roots = −1 + 5 + ( −1 − 5 ) = −2 Product of the roots = ( −1 + 5 )( −1 − 5 ) = ( −1)2 − ( 5 )2 = −4
= ( − p)2 − 3q = p 2 − 3q
(b) α 3β + αβ3 = αβ(α 2 + β 2 ) = αβ[(α + β)2 − 2αβ] = q[( − p)2 − 2 q] = p2 q − 2q 2
(c)
1 1 + 2α + β α + 2β α + 2β + 2α + β = (2α + β)(α + 2β) 3(α + β) = 2 2α + 4αβ + αβ + 2β 2 3(α + β) = 2 2(α + 2αβ + β 2 ) + αβ 3(α + β) = 2(α + β)2 + αβ 3( − p) = 2( − p ) 2 + q −3 p = 2 p2 + q
One required equation is x 2 + 2 x − 4 = 0 . −5 + 3 2 −5 − 3 2 + 2 2 −10 = 2 = −5
(d) Sum of the roots =
−5 + 3 2 −5 − 3 2 )( ) 2 2 ( −5)2 − (3 2 )2 = 4 7 = 4 7 One required equation is x 2 + 5 x + = 0 , 4 i.e. 4 x 2 + 20 x + 7 = 0 . Product of the roots = (
5. x 2 − 4 x + 1 = 0 α + β = 4 , αβ = 1 (a) Sum of the roots = 2α + 1 + 2β + 1 = 2(α + β) + 2 = 10
Chapter 1 Quadratic Equations & Quadratic Functions
Product of the roots = (2α + 1)(2β + 1) = 4αβ + 2(α + β) + 1 = 13 One required equation is x 2 − 10 x + 13 = 0 . 3 3 α+β + = 3( ) = 12 α β αβ 3 3 1 Product of the roots = ( )( ) = 9( ) = 9 α β αβ One required equation is x 2 − 12 x + 9 = 0 .
(b) Sum of the roots =
(c) Sum of the roots 1 1 = α 2 + + β2 + α β = (α + β)2 − 2αβ +
α+β αβ
= 18 Product of the roots 1 1 = (α 2 + )(β 2 + ) α β
= α 2β 2 + =2+
7. 8 x 2 − ax + 9 = 0 Let α, 2α be the roots. a ............(1) 8 9 Product of the roots = α(2α ) = 2α 2 = ......(2) 8 3 Solving (2), α = ± , 4 3 When α = , a = 18 , 4 3 When α = − , a = −18 4 3 3 ∴ When a = 18 , the roots are , . 4 2 3 3 When a = −18 , the roots are − , − . 4 2 Sum of the roots = α + 2α = 3α =
8. 12 x 2 + mx − 5 = 0 m 12 −5 Product of the roots = 12 m 5 11 − − (− ) = 12 12 6 m 5 11 − + = 12 12 6 m = −17
Sum of the roots = −
α 2 β2 1 + + β α αβ
α 3 + β3 αβ
= 2 + (α + β)(α 2 − αβ + β 2 ) = 2 + 4[(α + β)2 − 3αβ] = 2 + 4(16 − 3) = 54 One required equation is x 2 − 18 x + 54 = 0 .
The equation becomes 12 x 2 − 17 x − 5 = 0 ( 4 x + 1)(3 x − 5) = 0 x=−
(d) Sum of the roots
= α −β+β−α = 0
1 or 5 4 3
9. kx 2 + 4 x + k 2 − 21 = 0
Product of the roots
k 2 − 21 k =4
Product of the roots =
= (α − β)(β − α ) = αβ − α 2 − β 2 + αβ
k 2 − 4 k − 21 = 0 ( k − 7)( k + 3) = 0 k = 7 or −3
= −(α 2 − 2αβ + β 2 ) = −[(α + β)2 − 4αβ] = −12 One required equation is x 2 − 12 = 0 .
10. ( kx − 1)2 = k k 2 x 2 − 2 kx + 1 − k = 0
6. Let the other root be α. Sum of the roots = 2 − 5 + α = 4 α =2+ 5 ∴ The other root of the equation is 2 + 5 . Product of the roots = l = (2 − 5 )(2 + 5 ) = (2) 2 − ( 5 ) 2 = −1
23
1− k −3 = 16 k2 16 − 16k = −3k 2
Product of the roots =
3k 2 − 16k + 16 = 0 (3k − 4)( k − 4) = 0 4 k= or 4 3
24
Chapter 1 Quadratic Equations & Quadratic Functions
11. (a)
x 2 + a(3a − 5) x = 2( x + 4 a) x 2 + a(3a − 5) x − 2( x + 4 a) = 0 x 2 + a(3a − 5) x − 2 x − 8a = 0 x 2 + [a(3a − 5) − 2]x − 8a = 0...........(*) Let α, −α be the roots of (*). Sum of the roots = −[a(3a − 5) − 2] = α − α = 0 3a 2 − 5a − 2 = 0 ( a − 2)(3a + 1) = 0 1 a = 2 or − 3
Product of the roots 1 1 1 .....................(4) =( )( )= 2α + 1 2β + 1 35 By (4),
By (1) and (2),
(b) If a > 0 , by (a), a = 2 . The equation becomes x 2 + 2 x = 2( x + 8)
x 2 + 2 x = 2 x + 16 x 2 = 16 x = ±4 12. α and β are the roots of x 2 + px − 5 = 0 , Sum of the roots = α + β = − p .............(1) Product of the roots = αβ = −5 ............(2) α 2 and β 2 are the roots of x 2 − 19 x + q = 0 ,
1 1 )( )= 2α + 1 2β + 1 1 = (2α + 1)(2β + 1) 1 = 4αβ + 2(α + β) + 1 (
By (3),
1 35 1 35 1 35
1 1 = 4 k + 2(5) + 1 35 1 1 = 4 k + 11 35 4 k + 11 = 35 k=6
n 1 1 + =− 2α + 1 2β + 1 35 n 2β + 1 + 2α + 1 =− (2α + 1)(2β + 1) 35 n 2(α + β ) + 2 =− (2α + 1)(2β + 1) 35
By (1) and (4),
2(5) + 2 n =− 35 35 n = −12
Sum of the roots = α 2 + β 2 = 19 ..........(3)
14. Let α and α + 1 be the roots of x 2 − px + q = 0 .
Product of the roots = α 2β 2 = q ..........(4)
Sum of the roots = α + α + 1 = 2α + 1 = p .....(1)
By (3), α 2 + β 2 = 19 (α + β)2 − 2αβ = 19 By (1) and (2), ( − p)2 − 2( −5) = 19
p 2 + 10 = 19 p2 = 9 p = ±3 By (2) and (4), ( −5)2 = q q = 25 13. α and β are roots of x 2 − 5 x + k = 0 , Sum of the roots = α + β = 5 .............(1) Product of the roots = αβ = k ...........(2) 1 1 are roots of 35 x 2 + nx + 1 = 0 , and 2β + 1 2α + 1 Sum of the roots 1 1 n = + =− ...................(3) 2α + 1 2β + 1 35
Product of the roots = α(α + 1) = q ..............(2) 1 By (1), α = ( p − 1) 2 1 Put α = ( p − 1) into (2), 2 1 1 ( p − 1)[ ( p − 1) + 1] = q 2 2 1 1 ( p − 1) ( p + 1) = q 2 2 p 2 − 1 = 4q
p2 − 4q − 1 = 0 15. Let α and α − 2 be the roots of x 2 + px + q = 0 . Sum of the roots = α + α − 2 = 2α − 2 = − p .................(1) Product of the roots = α(α − 2) = q ..................................(2) 1 By (1), α = (2 − p) 2 1 Put α = (2 − p) into (2), 2
Chapter 1 Quadratic Equations & Quadratic Functions
18. (a) Given α and β are roots of ax 2 + bx + c = 0 . b Sum of the roots = α + β = − a c Product of the roots = αβ = a ∴ α 2 + β 2 = (α + β)2 − 2αβ
1 p (2 − p)(1 − − 2) = q 2 2 1 p (2 − p)( − − 1) = q 2 2 1 1 [ −( p − 2)][− ( p + 2)] = q 2 2 1 ( p − 2)( p + 2) = q 4 p 2 − 4 = 4q
16. Let α and β be the roots of ax 2 + bx + c = 0 . b ....................(1) a c Product of the roots = αβ = ....................(2) a α m m = , ∴ α = β ..............................(3) Q β n n Sum of the roots = α + β = −
m b β+β = − n a m+n b ( )β = − n a β=−
Put (3) into (2), (
b 2 2c − a a2 1 2 = 2 (b − 2 ac) a c2 α 2β 2 = 2 a A quadratic equation with roots α 2 and β 2 is =
p 2 = 4 + 4q
Put (3) into (1),
25
bn a( m + n )
m c β)β = n a cn β2 = am b2n2 cn = 2 a ( m + n)2 am ab 2 mn 2 = a 2 cn( m + n)2 a 2 cn( m + n)2 an = ( m + n)2 ac
mnb 2 =
17. Let α and β be the roots of ax 2 + bx + c = 0 . b ....................(1) a c Product of the roots = αβ = ....................(2) a
Sum of the roots = α + β = −
α 2 + β 2 = 4 .................................................(3) By (3), α 2 + 2αβ + β 2 − 2αβ = 4
(α + β)2 − 2αβ = 4 b2 c − 2( ) = 4 2 a a b 2 − 2 ac = 4 a 2 b 2 = 2 ac + 4 a 2
c2 1 x 2 − 2 (b 2 − 2 ac) x + 2 = 0 a a a 2 x 2 − (b 2 − 2 ac) x + c 2 = 0
(b) α and β are roots of 2 x 2 − 4 x − 1 = 0 . By (a), take a = 2 , b = −4 , c = −1 . A quadratic equation with roots α 2 and β 2 is
4 x 2 − [16 − 2(2)( −1)]x + ( −1)2 = 0 4 x 2 − 20 x + 1 = 0 By (a), take a = 4 , b = −20 , c = 1. A quadratic equation with roots α 4 and β 4 is
16 x 2 − [20 2 − 2( 4)(1)]x + 1 = 0 16 x 2 − 392 x + 1 = 0 19. (a) α and β are the roots of 2 x 2 + ax + b = 0 . a Sum of the roots = α + β = − 2 b Product of the roots = αβ = 2 (b) (α − 1) and (β − 1) are the roots of 2 x 2 + mx + n = 0 . Sum of the roots m = (α − 1) + (β − 1) = − ...............(1) 2 Product of the roots n = (α − 1)(β − 1) = ......................(2) 2 m By (1), α + β − 2 = − 2 a m By (a), − − 2 = − 2 2 a m +2= 2 2 m=a+4
26
Chapter 1 Quadratic Equations & Quadratic Functions
n 2 n αβ − (α + β) + 1 = 2 b a n By (a), − ( − ) + 1 = 2 2 2 b+a+2=n n=a+b+2 By (2),
(α − 1)(β − 1) =
By (2),
(c) (i) α and β are the roots of 4 x 2 + 3 x − 5 = 0 . 3 4 5 Product of the roots = αβ = − 4 (α − 1) and (β − 1) are the roots of Sum of the roots = α + β = −
bñÉêÅáëÉ=Na=EéKOUF 1. (a)
(α − 1) + (β − 1) = −
∴
(b) For all real values of x,
( x − 1)2 ≥ 0 4( x − 1)2 ≥ 0 4( x − 1)2 + 3 ≥ 0 + 3 4 x 2 − 8x + 7 ≥ 3
d (α − 1)(β − 1) = 4 d αβ − (α + β) + 1 = 4 5 3 d − − (− ) + 1 = 4 4 4 d=2
∴ The minimum value of 4 x 2 − 8 x + 7 is 3 . 2. (a)
(ii) α and β are the roots of 2 x 2 − 3 x − 1 = 0 . Sum of the roots = α + β =
4 x 2 − 8x + 7 7 = 4( x 2 − 2 x + ) 4 2 7 2 2 = 4[ x − 2 x + ( )2 − ( )2 + ] 2 2 4 = 4( x − 1)2 + 3 ∴ a = −1 , b = 3
4 x 2 + cx + d = 0 .
c 4 c (α + β ) − 2 = − 4 3 c − −2= − 4 4 c = 11
f 2 f αβ − 2(α + β) + 4 = 2 1 3 f − − 2( ) + 4 = 2 2 2 f =1 (α − 2)(β − 2) =
3 2
1 2 (α − 2) and (β − 2) are the roots of
Product of the roots = αβ = −
2 x 2 + ex + f = 0 . Sum of the roots = (α − 2) + (β − 2) e = − ...................(1) 2 Product of the roots = (α − 2)(β − 2) f = ................(2) 2 e By (1), α + β − 4 = − 2 3 e −4=− 2 2 e=5
3 x 2 − 12 x + 14 14 = 3( x 2 − 4 x + ) 3 4 4 14 2 = 3[ x − 4 x + ( )2 − ( )2 + ] 2 2 3 = 3( x − 2)2 + 2 ∴ p = −2 , q = 2
(b) For all real values of x, ( x − 2)2 ≥ 0 3( x − 2)2 ≥ 0 3( x − 2)2 + 2 ≥ 0 + 2 3 x 2 − 12 x + 14 ≥ 2 1 1 ≤ 3 x 2 − 12 x + 14 2 8 8 ≤ 2 2 3 x − 12 x + 14 8 ≤4 2 3 x − 12 x + 14 ∴ The maximum value of is 4 .
8 3 x 2 − 12 x + 14
Chapter 1 Quadratic Equations & Quadratic Functions
3. (a)
3 x 2 − 6 x + 10 10 = 3( x 2 − 2 x + ) 3 2 2 10 2 = 3[ x − 2 x + ( )2 − ( )2 + ] 2 2 3 2 = 3( x − 1) + 7
(b) For all real values of x, ( x − 1)2 ≥ 0 3( x − 1)2 ≥ 0 3( x − 1)2 + 7 ≥ 7 1 1 ≤ 2 3( x − 1) + 7 7 5 5 ≤ 2 3 x − 6 x + 10 7 5 5 − 2 ≥− 7 3 x − 6 x + 10 5 5 ≥1− 1− 2 7 3 x − 6 x + 10 2 = 7 ∴ The given expression is always greater 2 than or equal to . 7 2 7 x+ ) 3 3 2 2 2 7 = −3[ x 2 − x + ( )2 − ( )2 + ] 3 6 6 3 1 20 = −3( x − )2 − 3 3 For all real values of x, 1 ( x − )2 ≥ 0 3 1 2 −3( x − ) ≤ 0 3 1 2 20 −20 −3( x − ) − ≤ 3 3 3 20 2 −3 x + 2 x − 7 ≤ − 3 ∴ −3 x 2 + 2 x − 7 is always negative for all real values of x.
4. −3 x 2 + 2 x − 7 = −3( x 2 −
5. The graph of y = ax 2 − 8 x − 4 lies below the x-axis for all values of x, i.e. a < 0 and D < 0 . D = ( −8)2 − 4 a( −4 ) < 0 64 + 16 a < 0 a < −4 6. Consider 2 x 2 + 2 ax − (b + 1) = 0 .............. (∗1 ) Let α and β be the roots of (∗1 ) . 2a Sum of the roots = α + β = − = −a 2
Product of the roots = αβ = −
27
(b + 1) 2
b−a x − (b − 1) = 0 .......... (∗2 ) 2 Since (∗1 ) and (∗2 ) cut the same points on the points on the x-axis, α and β are also the roots of (∗2 ) . b−a ) Sum of the roots = α + β = −( 2 Product of the roots = αβ = −(b − 1) (b + 1) − = −(b − 1) 2 b + 1 = 2b − 2 b=3 b−a − a = −( ) 2 2a = b − a 3a = b = 3 a =1 Consider x 2 +
7. (a) f ( x ) = − x 2 + qx + r = −( x 2 − qx − r ) q q = −[ x 2 − qx + ( )2 − ( )2 − r ] 2 2 q q2 = −( x − ) 2 + ( + r) 2 4 (b) The maximum value of f ( x ) occurs when x = 3. q ∴ 3− = 0 2 q=6 (c) f ( x ) ≤ 0 q q2 −( x − ) 2 + ( + r) ≤ 0 2 4 Form (b),
q2 +r =0 4 62 +r =0 4 r = −9
8. (a) y-intercept = k x-intercept = 1, 3 k f ( x ) = ( x − 1)( x − 3) 3 k ( x − 1)( x − 3) 3 k = ( x 2 − 4 x + 3) 3 k 2 4 4 = [ x − 4 x + ( )2 − ( )2 + 3] 3 2 2 k k = ( x − 2)2 − 3 3
(b) f ( x ) =
28
Chapter 1 Quadratic Equations & Quadratic Functions
For all real values ( x − 2)2 k ( x − 2)2 3 k k ( x − 2)2 − 3 3 k ∴ f ( x) ≥ − 3
of x, ≥0
(α + β)2 − 2αβ = 9 4a + 1 ( −4)2 − 2( )=9 a 8a + 2 16 − =9 a −a − 2 = 0 a = −2
≥0 ≥0−
α 2 + β2 = 9
If
k 3
k ∴ The least value of f ( x ) is − . 3 (c) If f ( x ) ≥ −2 , k from (b), − = −2 3 k=6 k f ( x ) = ( x − 1)( x − 3) 3 6 = ( x − 1)( x − 3) 3 = 2( x − 1)( x − 3)
y
(c) y = −2(x + 2)2 + 1 −4 − 2 2
1 −2 −1 −4 + 2 2
x
O
−2
−4
= 2 x − 8x + 6 2
y = x + m........................(1) 9. (a) 2 y = x + 4 x − 3m...........(2) P( x1, y1 ) a n d Q( x2 , y2 ) a r e p o i n t s o f intersection, By (1) and (2), x + m = x 2 + 4 x − 3m x 2 + 3x − 4m = 0
(b) If P and Q coincide, x 2 + 3 x − 4 m = 0 has two equal real roots, i.e. D = 0 . 32 − 4( −4 m) = 0 9 m=− 16 10. (a) Consider y = a( x + 2) + 1 . 2
y = a( x 2 + 4 x + 4 ) + 1 y = ax 2 + 4 ax + 4 a + 1 D = ( 4 a)2 − 4( a)( 4 a + 1)
= 16 a 2 − 16 a 2 − 4 a = −4 a If a < 0 , D > 0 ∴ The graph of y = a( x + 2)2 + 1 cuts the x-axis at two distinct points if a < 0 . (b) From (a), y = ax 2 + 4 ax + 4 a + 1......(∗) α and β are roots of ax 2 + 4 ax + 4 a + 1 = 0 . 4a Sum of the roots = α + β = − = −4 a 4a + 1 Product of the roots = αβ = a
−6
bñÉêÅáëÉ=Nb=EéKPQF 1. − 2 + 7 − 9 − 15 − − 3 = 5 − −6 − −3 = 5−6−3 = −4 2. 4 − 9 + 7 − 16 − 12 − 5 = −5 + −9 − 7 = 5+9−7 =7 3. f ( x ) = 2 x − 5 + x f ( 0 ) = 2( 0 ) − 5 + 0 = − 5 = 5 f (1) = 2(1) − 5 + 1 = − 3 + 1 = 3 + 1 = 4 f ( −3) = 2( −3) − 5 + − 3 = − 11 + − 3 = 11 + 3 = 14
4. g( x ) = g( −6) =
x x −6 −6 = = −1 −6 6
Chapter 1 Quadratic Equations & Quadratic Functions
−3 −3 = = −1 −3 3 7 7 g(7) = = =1 7 7 g( −3) =
12. x 2 − 5 x + 5 = 1
5. 3 x + 7 = 5 3x + 7 = 5
or 2 x=− or 3
3 x + 7 = −5 x = −4
x + 1 = 3x − 2 x + 1 = 3 x − 2 or x + 1 = −(3 x − 2) 3 = 2x or x + 1 = −3 x + 2 3 or x= 4x = 1 2 3 1 or x= x= 2 4
or x + 1 = −3 x + 2 x + 1 = 3x − 2 3 1 or x= x= 2 4 But 3 x − 2 should be non-negative, 1 1 Q 3( ) − 2 < 0 , ∴ x = is not a solution. 4 4 3 3 Q 3( ) − 3 > 0 , ∴ x = 2 2
10.
x 2 − 5 x + 5 = −1
x 2 − 5x + 4 = 0 ( x − 1)( x − 4) = 0 x = 1, 4 ∴ x = 1, 2, 3, 4
or or or
x 2 − 5x + 6 = 0 ( x − 2)( x − 3) = 0 x = 2, 3
x 2 − x − 4 = −2 or or x2 − x − 2 = 0 x2 − x − 6 = 0 or ( x − 2)( x + 1) = 0 ( x − 3)( x + 2) = 0 x = 2, − 1 x = 3, − 2 or ∴ x = −1, − 2, 2, 3 14. (2 x − 1)2 + 2 x − 1 − 12 = 0 2
2 x − 1 + 2 x − 1 − 12 = 0 ( 2 x − 1 − 3)( 2 x − 1 + 4) = 0
2x − 1 = 3 2x − 1 = 3 x=2
8. x + 1 = 3 x − 2
2x − 1 =9 5 2 x − 1 = 45 2 x − 1 = 45 2 x = 46 x = 23
or
x2 − x − 4 = 2
Since any absolute value should be non-negative, ∴ The equation has no solution.
9.
x 2 − 5x + 5 = 1
13. x 2 − x − 4 = 2
6. − 3 x = −6
7.
29
2 x − 1 = −4 (rejected) 2 x − 1 = −3 x = −1
or or or
15. 2( x + 3)2 − 3 x + 3 − 14 = 0 2
2 x + 3 − 3 x + 3 − 14 = 0 (2 x + 3 − 7)( x + 3 + 2) = 0 2 x + 3 = 7 or x + 3 = −2 (rejected) 2( x + 3) = 7 or 2( x + 3) = −7 1 13 or x= x=− 2 2 16 − 17. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
2 x − 1 = −45 2 x = −44 x = −22
or or or
6−x 1 = 3+ x 2 6−x 1 = 3+ x 2 12 − 2 x = 3 + x 9 = 3x x=3
y
18.
y = x −1 2 1
or or or or
6−x 1 =− 3+ x 2 12 − 2 x = −3 − x 15 = x x = 15
3( x + 4) = −4 x + 6 3 x + 12 = −4 x + 6 6 or x = 18 x=− 7 For the same reason as question 8, x = 18 .
1
3
y
19.
11. 3 x + 4 = 4 x − 6 3( x + 4) = 4 x − 6 or 3 x + 12 = 4 x − 6 or
x
O
1 −2
−1 O −1
y = − x +1 x 1
2
3
30
Chapter 1 Quadratic Equations & Quadratic Functions
y
20.
Product of the roots = αβ = − (a) α 2β + αβ 2 = αβ(α + β) 5 = −2( ) 2 = −5
6 5 4
(b) (α − 3β)(3α − β) = 3α 2 − αβ − 9αβ + 3β 2
2
y= x +x −6
3
= 3(α 2 + β 2 ) − 10αβ = 3[(α + β)2 − 2αβ] − 10αβ 5 = 3[( )2 − 2( −2)] − 10( −2) 2 3 = 50 4
2 1
−3
−2
−1
O
4 = −2 2
x 1
2
3
6. α and β are roots of x 2 − 2 x − 1 = 0 . Sum of the roots = α + β = 2 Product of the roots = αβ = −1 (a) α + 2β + β + 2α = 3α + 3β = 3(α + β) = 3(2) =6 (α + 2β)(β + 2α ) = 2α 2 + 5αβ + 2β 2
oÉîáëáçå=bñÉêÅáëÉ=N=EéKPSF 2(3 x 2 + 5) = 19 x
1.
6 x − 19 x + 10 = 0 (3 x − 2)(2 x − 5) = 0 5 2 x= or 3 2 2
2.
4( x − 1)2 + 5( x − 1) − 6 = 0 [ 4( x − 1) − 3][( x − 1) + 2] = 0 ( 4 x − 7)( x + 1) = 0 7 x= 4
= 2(α 2 + β 2 ) + 5αβ = 2[(α + β)2 − 2αβ] + 5αβ
or −1
3. The equation x 2 − 2 kx + 36 = 0 has equal roots. D = ( −2 k )2 − 4(1)(36) = 0
4 k 2 − 144 = 0 k 2 = 36 k = ±6 4. The equation (6k + 1) x 2 − 2(2 k − 3) x + 1 = 0 has equal roots.
D = [ −2(2 k − 3)]2 − 4(6k + 1) = 0 4( 4 k 2 − 12 k + 9) − 4(6k + 1) = 0
= 2[(2)2 − 2( −1)] + 5( −1) =7 One required equation is x 2 − 6 x + 7 = 0 . (b) α 2β + αβ 2 = αβ(α + β) = −1(2) = −2 (α 2β)(αβ 2 ) = α 3β3 = (αβ)3 = ( −1)3 = −1 One required equation is x 2 + 2 x − 1 = 0 .
7. y = 3 x 2 + 12 x + 7 + k ( x 2 − 1) = 3 x 2 + 12 x + 7 + kx 2 − k
4 k − 12 k + 9 − 6k − 1 = 0
= (3 + k ) x 2 + 12 x + 7 − k D = 12 2 − 4(7 − k )(3 + k )
4 k 2 − 18k + 8 = 0
= 144 − 4(21 + 4 k − k 2 )
2
2 k 2 − 9k + 4 = 0 (2 k − 1)( k − 4) = 0 1 or 4 k= 2 5. α and β are roots of 2 x 2 − 5 x − 4 = 0 . 5 Sum of the roots = α + β = 2
= 144 − 84 − 16k + 4 k 2 = 4 k 2 − 16k + 60 = 4( k 2 − 4 k + 15) 4 4 = 4[k 2 − 4 k + ( )2 − ( )2 + 15] 2 2 = 4( k − 2)2 + 44
Chapter 1 Quadratic Equations & Quadratic Functions
Q ( k − 2)2 ≥ 0 , ∴ D > 0 ∴ The graph of y = 3 x 2 + 12 x + 7 + k ( x 2 − 1) cuts the x-axis for any value of k. 8. α and β are roots of 2 x − 3 x − 5 = 0 . 3 Sum of the roots = α + β = 2 5 Product of the roots = αβ = − 2 α + 2 and β + 2 are roots of 2 x 2 + px + q = 0 . p Sum of the roots = (α + 2) + (β + 2) = − .......(1) 2 q Product of the roots = (α + 2)(β + 2) = ........(2) 2 p By (1), α + β + 4 = − 2 3 p +4=− 2 2 p = −11 q By (2), (α + 2)(β + 2) = 2 q αβ + 2(α + β) + 4 = 2 5 3 q − + 2( ) + 4 = 2 2 2 q=9
(b) When k = −5 , the equation becomes x 2 + [( −5)2 + 14( −5) + 45]x + 2( −5) − 5 = 0 x 2 − 15 = 0 x = ± 15
2
When k = −9 , the equation becomes x 2 + [( −9)2 + 14( −9) + 45]x + 2( −9) − 5 = 0 x 2 − 23 = 0 x = ± 23 11. (a) x 2 − ( k + 3) x + ( k 2 − 3k + 6) = 0
D = [ −( k + 3)]2 − 4( k 2 − 3k + 6) = 0 k 2 + 6k + 9 − 4 k 2 + 12 k − 24 = 0 −3k 2 + 18k − 15 = 0 k 2 − 6k + 5 = 0 ( k − 5)( k − 1) = 0 k = 1 or 5 (b) 2 kx 2 + 4( 4 k − 5) x + (2 k + 5) = 0 D = [ 4( 4 k − 5)]2 − 4(2 k )(2 k + 5) = 0
16(16k 2 − 40 k + 25) − (16k 2 + 40 k ) = 0 256k 2 − 640 k + 400 − 16k 2 − 40 k = 0 240 k 2 − 680 k + 400 = 0 6k 2 − 17k + 10 = 0 (6k − 5)( k − 2) = 0 5 k = or 2 6
9. (a) α 3 + β3 = (α + β)(α 2 − αβ + β 2 ) = 335 5(α 2 − αβ + β 2 ) = 335 α 2 + β 2 − αβ = 67......(∗) (α + β)2 − 2αβ − αβ = 67 52 − 3αβ = 67 αβ = −14
By (∗) , α 2 + β 2 = 67 + αβ = 67 − 14 = 53 (b) α 2 and β 2 are roots of equation Sum of the roots = α 2 + β 2 = 53 Product of the roots = α 2β 2 = (αβ)2 = ( −14)2 = 196 One required equation is x 2 − 53 x + 196 = 0 .
10. (a) Let α, −α be the roots of x 2 + ( k 2 + 14 k + 45) x + 2 k − 5 = 0 . Sum of the roots = α + ( −α ) =0
= −( k 2 + 14 k + 45) k + 14 k + 45 = 0 ( k + 5)( k + 9) = 0 k = −5 2
12 − 16. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons. 17. α and β are roots of ax 2 + bx + c = 0 . b Sum of the roots = α + β = − a c Product of the roots = αβ = a (a) α 3 + β3 = (α + β)(α 2 − αβ + β 2 ) = (α + β)[(α + β)2 − 2αβ − αβ] = (α + β)[(α + β)2 − 3αβ] b −b c = − [( )2 − 3( )] a a a b b 2 − 3ac ) =− ( a a2 − b(b 2 − 3ac) = a3 α 3β3 = (αβ)3 =
or −9
31
c3 a3
32
Chapter 1 Quadratic Equations & Quadratic Functions
If x = β , f [ f (β)] − β = f (β) − β =β−β =0 ∴ α, β are also the roots of f [ f ( x )] − x = 0 .
A quadratic equation with roots α 3 and β3 is b(b 2 − 3ac) c3 x + 3 = 0, 3 a a 3 2 2 i.e. a x + b(b − 3ac) x + c 3 = 0 x2 +
(b) By (a), take a = 1 , b = −3 , c = −2 . One required equation is (1) x 2 + ( −3)[( −3)2 − 3(1)( −2)]x + ( −2)3 = 0 , i.e. x 2 − 45 x − 8 = 0
(b) For f ( x ) = x 2 − 3 x + 2 , We set f ( x ) − x = 0 , i.e. x 2 − 4 x + 2 = 0 −( −4) ± ( −4)2 − 4(2) 2 4± 8 = 2 =2± 2
x=
18. α and β are roots of x 2 + ax + b = 0 . Sum of the roots = α + β = − a Product of the roots = αβ = b
By (a), f [ f ( x )] − x = 0 has roots 2 ± 2 . f [ f ( x )] − x = 0
(a) (i) α 3 + β3 = (α + β)(α 2 − αβ + β 2 ) = (α + β)[(α + β)2 − 2αβ − αβ]
( x 2 − 3 x + 2)2 − 3( x 2 − 3 x + 2) + 2 − x = 0
= − a[( − a)2 − 3b]
x 4 − 6 x 3 + 13 x 2 − 12 x + 4 − 3 x 2 + 9 x − 6+2−x =0
= − a( a 2 − 3b) (ii) (α − β 2 + 1)(β − α 2 + 1) = αβ − α + α − β + α β − β 3
3
2 2
x 4 − 6 x 3 + 10 x 2 − 4 x = 0 2
x ( x 3 − 6 x 2 + 10 x − 4) = 0
+ β − α2 + 1 = αβ − (α 3 + β3 ) + (α + β)
∴
x ( x − 2)( x 2 − 4 x + 2) = 0 x = 0, 2 or 2 ± 2
− (α 2 + β 2 ) + (αβ)2 + 1
20. (a) Let α be the common root of the two equations
= αβ − (α 3 + β3 ) + (α + β)
3α 2 + aα + b = 0 ..........(1) 2 3α + bα + a = 0 .........(2)
− [(α + β) − 2αβ] + (αβ) + 1 2
2
= b + a( a 2 − 3b) − a − ( a 2 − 2 b) + b 2 + 1 = b + a 3 − 3ab − a − a 2 + 2 b + b 2 + 1
(1) − (2), ( a − b)α = a − b α = 1 (Q a ≠ b)
= b 2 + 3b − 3ab + a 3 − a 2 − a + 1 = b 2 − 3( a − 1)b + a 3 + 1 − ( a 2 + a) = b − 3( a − 1)b + ( a + 1)( a − a + 1) − a( a + 1) 2
2
= b 2 − 3( a − 1)b + ( a + 1)( a 2 − 2 a + 1) = b 2 − 3( a − 1)b + ( a − 1)2 ( a + 1)
(b) If one root of the equation plus 1 is equal to the square of the other, i.e. α + 1 = β 2 or β + 1 = α 2 , then (α + 1 − β 2 )(β + 1 − α 2 ) = 0 . From (a)(ii), b 2 − 3( a − 1)b + ( a − 1)2 ( a + 1) = 0 19. f ( x ) = ax 2 + bx + c (a) α and β are roots of f ( x ) − x = 0 . f (α ) = α ∴ f (α ) − α = 0 f (β) − β = 0 f (β) = β For the equation, f [ f ( x )] − x = 0 If x = α , f [ f (α )] − α = f (α ) − α =α−α =0
(b) As α = 1, 3 + a + b = 0 a + b = −3 a and b are roots of x 2 + hx + k = 0 Sum of the roots = a + b = − h −3 = − h h =3 Product of the roots = ab = k Since h, k are positive integers and a + b = −3 , k = ab = ( −1)( −2) = 2 is the only solution. ∴ k=2 21. No solution is provided for the H.K.C.E.E. question because of the copyright reasons.
båêáÅÜãÉåí=N=EéKPUF 1. (a)
x 2 − 6 acx + a 2 (9c 2 − 4b 2 ) = 0 x 2 − 6 acx + [a(3c + 2 b) ⋅ a(3c − 2 b)] = 0 [ x − a(3c + 2 b)][ x − a(3c − 2 b)] = 0 x = a(3c ± 2 b)
33
Chapter 1 Quadratic Equations & Quadratic Functions
Since kf ( x ) = g( x ) has equal roots, D = [ −(6k + 3)]2 − 4 k[ −(3 + 3k )] = 0
(1 − l 2 ) x 2 − 2 mx + m 2 = 0
(b)
(1 + l )(1 − l ) x − 2 mx + m = 0 [(1 + l ) x − m][(1 − l ) x − m] = 0 2
2
x= 4
36k 2 + 36k + 9 + 12 k + 12 k 2 = 0
m m or 1− l 1+ l
48k 2 + 48k + 9 = 0 16k 2 + 16k + 3 = 0 ( 4 k + 3)( 4 k + 1) = 0
2
x 3 − 5x 3 + 4 = 0
2. 2
k =−
2
( x 3 − 4)( x 3 − 1) = 0 2
x3 = 4 x = ±8
or or
2
x3 =1 x = ±1
22 x +2 = 9 ⋅ 2 x − 2
3.
22 x ⋅ 22 = 9 ⋅ 2 x − 2 4⋅2
2x
− 9 ⋅ 2x + 2 = 0
7. ax 2 + bx + c = 0 and px 2 + qx + r = 0 have one root in common. Let the common root be α. aα 2 + bα + c = 0 .......(1) 2 pα + qα + r = 0 ......(2) (1) × p − (2) × a, bpα + cp − aqα − ar = 0 α(bp − aq ) = ar − cp ar − cp α= bp − aq (1) × q − (2) × b,
(2 x − 2)( 4 ⋅ 2 x − 1) = 0 1 2 x = 2 or 2 x = 4 x = 1 or x = −2
x + 9 + 11 = x x + 9 = x − 11
4.
aqα 2 + cq − bpα 2 − br = 0
x + 9 = x 2 − 22 x + 121 x 2 − 23 x + 112 = 0 ( x − 16)( x − 7) = 0 x = 16 or 7 Check: When x = 16 ,
∴
L.H.S. = 16 + 9 + 11 = 16 = R.H.S. When x = 7 ,
∴ 5.
L.H.S. = 7 + 9 + 11 = 15 ≠ R.H.S. x = 16
1 3 or − 4 4
α 2 ( aq − bp) = br − cq br − cq α2 = aq − bp ar − cp 2 br − cq ( ) = bp − aq aq − bp 2 ( ar − cp) = br − cq aq − bp (br − cq )( aq − bp) = ( ar − cp)2 y
8.
y= x ⋅ x 8 6
(1 − x ) + (1 − x ) − 6 = 0 [(1 − x ) − 2][(1 − x ) + 3] = 0 or 1 − x = −3 1− x = 2 x =4 x = −1 (rejected) or 2
∴
x = 16
3(1 + k ) 6. f ( x ) = x − 6 x − , g( x ) = 3 x k kf ( x ) = g( x )
4 2
−3
−2
−1
O −2
2
kx 2 − 6kx − 3(1 + k ) = 3 x kx 2 − 6kx − 3 − 3k − 3 x = 0 kx 2 − (6k + 3) x − (3 + 3k ) = 0
−4 −6 −8
x 1
2
3
34
Chapter 1 Quadratic Equations & Quadratic Functions
`ä~ëëïçêâ=N=EéKPF
4 ( x − )2 − ( 3 4 13 4 (x − + )( x − − 3 3 3
1. 3 x − 14 x + 8 = 0 ( x − 4)(3 x − 2) = 0 2 x = 4 or 3 2
2.
6 x 2 − 13 x − 5 = 0 (2 x − 5)(3 x + 1) = 0 5 x= 2
1 3
or −
1
1 4
( x + 1)2 − ( 1 + m )2 = 0 ( x + 1 − 1 + m )( x + 1 + 1 + m ) = 0 x = −1 ± 1 + m
1
y 2 − 3y 4 + 2 = 0
4.
x2 + 2x − m = 0 2 2 x 2 + 2 x + ( )2 − ( )2 − m = 0 2 2 ( x + 1)2 − (1 + m) = 0
4.
3. ( y − 3)2 − 10( y − 3) − 56 = 0 [( y − 3) + 4][( y − 3) − 14] = 0 y − 3 = −4 or y − 3 = 14 y = −1 or y = 17
13 2 ) =0 3 13 )=0 3 4 ± 13 x= 3
( y − 2)( y − 1) = 0 1
4n 2 4n ) − ( )2 + 1 2 2 = ( x − 2 n) 2 − 4 n 2 + 1
5. x 2 − 4n + 1 = x 2 − 4n + (
1 4
1
y 4 = 2 or y 4 = 1 y = 16 or y = 1
= ( x − 2 n) 2 + 1 − 4 n 2 ∴
a = −2 n , b = 1 − 4n 2
`ä~ëëïçêâ=O=EéKSF 1.
x2 + 4x + 2 = 0 4 4 x 2 + 4 x + ( )2 − ( )2 + 2 = 0 2 2 ( x + 2)2 − 2 = 0
`ä~ëëïçêâ=P=EéKUF 1. Let a = 2 , b = −7 , c = 4 . Using the formula,
( x + 2)2 − ( 2 )2 = 0 ( x + 2 − 2 )( x + 2 + 2 ) = 0 x = −2 ± 2 2.
3.
x 2 − 3x + 1 = 0 3 3 x 2 − 3 x + ( )2 − ( )2 + 1 = 0 2 2 3 2 5 (x − ) − = 0 2 4 3 2 5 2 (x − ) − ( ) = 0 2 2 3 5 3 5 (x − + )( x − − )=0 2 2 2 2 3± 5 x= 2
3x 2 − 8 x + 1 = 0 8 1 x2 − x + = 0 3 3 8 2 8 2 1 2 8 x − x +( ) −( ) + = 0 3 6 6 3 4 2 13 (x − ) − =0 3 9
x=
2.
−( −7) ± ( −7)2 − 4(2)( 4) 7 ± 17 = 2( 2 ) 4
9 + 3x − 4 x 2 = 0 −4 x 2 + 3 x + 9 = 0 Let a = −4 , b = 3 , c = 9 . Using the formula, x=
−3 ± 32 − 4( −4)(9) −3 ± 153 3 ± 3 17 = = 2( −4) −8 8
5 x + 12 = 3 x 2
3.
3 x 2 − 5 x − 12 = 0 Let a = 3 , b = −5 , c = −12 . Using the formula, −( −5) ± ( −5)2 − 4(3)( −12) 2(3) 5 ± 169 x= 6 4 x = 3 or x = − 3 x=
Chapter 1 Quadratic Equations & Quadratic Functions
4. Let a = 3 , b = 4 , c = 5 . Using the formula,
−4 ± 4 − 4(3)(5) −4 ± −44 = 2(3) 6 The equation has no real solution. x=
2
5. Since 4 is a root of 3 x 2 + ax + 8 = 0 . ∴ 3( 4)2 + a( 4) + 8 = 0 4 a = −56 a = −14 Substitute a = −14 into the given equation, 3 x 2 − 14 x + 8 = 0 ( x − 4)(3 x − 2) = 0 2 3 2 ∴ The other root of the equation is . 3 ∴
x = 4 or
`ä~ëëïçêâ=Q=EéKVF 5x 2 + 9 x − 2 = 0
D = 9 2 − 4(5)( −2) = 121 = 112 D is a perfect square.
35
`ä~ëëïçêâ=R=EéKNNF 1. 2 x 2 − x − 21 = 0 D = ( −1)2 − 4(2)( −21) = 169
= 132 Q D > 0 and D is a perfect square. ∴ The equation has two rational roots. 2. 9 − 5 x + x 2 = 0 x 2 − 5x + 9 = 0
D = ( −5)2 − 4(1)(9) = −11 Q D<0 ∴ The equation has unreal roots. 3. x 2 − 2 x = 5
x2 − 2x − 5 = 0 D = ( −2)2 − 4( −5) = 24 Q D > 0 and D is not a perfect square. ∴ The equation has two real roots. 4. 4 x 2 − 28 x + 49 = 0 D = ( −28)2 − 4( 4)( 49) =0 Q D=0 ∴ The equation has two equal rational roots.
x 2 + 8 x + 16 = 0
D = 82 − 4(16) =0
D is a perfect square. 2 x − 7x + 3 = 0 2
D = ( −7)2 − 4(2)(3) = 25 = 52 D is a perfect square. 2 x 2 − 5x + 4 = 0
D = ( −5)2 − 4(2)( 4) = −7 D is not a perfect square.
9 x 2 − 30 x + 25 = 0 D = ( −30)2 − 4(9)(25) =0 D is a perfect square.
`ä~ëëïçêâ=S=EéKNNF 1. x 2 + 8 x + 5k = 0
D = 82 − 4(1)(5k ) = 64 − 20 k 2. 2 x 2 − kx − ( k − 3) = 0
D = ( − k )2 − 4(2)[ −( k − 3)] = k 2 + 8( k − 3) = k 2 + 8k − 24
`ä~ëëïçêâ=T=EéKNPF 1. x 2 − (2 m − 1) x − 2 m = 0 D = [ −(2 m − 1)]2 − 4( −2 m) = 4 m 2 − 4 m + 1 + 8m = 4m 2 + 4m + 1 = (2 m + 1)2 which is the square of a rational number. Therefore the roots of the given equation are rational.
36
Chapter 1 Quadratic Equations & Quadratic Functions
`ä~ëëïçêâ=V=EéKNSF
2. qx 2 + ( p + 3q ) x + 2 p = 0
1. 2 x 2 + 3 x − 7 = 0
D = ( p + 3q )2 − 4(q )(2 p) = p + 6 pq + 9q − 8 pq 2
2
= p 2 − 2 pq + 9q 2 = p 2 − 2 pq + q 2 + 8q 2 = ( p − q ) 2 + 8q 2 Q ( p − q )2 ≥ 0 and q 2 > 0 ∴ D>0 ∴ The equation has real roots.
3. (a) x 2 − kx + ( k + 3) = 0 ............(*) D = ( − k )2 − 4( k + 3)
3 2 7 Product of the roots = αβ = − 2 (a) α 2 + β 2 = (α + β)2 − 2αβ 3 7 = ( − ) 2 − 2( − ) 2 2 37 = 4
Sum of the roots = α + β = −
(b)
37
= k 2 − 4 k − 12 (b) If (*) has equal roots, D = 0. From (a), ( k + 2)( k − 6) = 0 k = −2 or 6 (c) For k = −2 , (*) becomes
x2 + 2x + 1 = 0
(c)
= 47 −2 37 =− 14 (α + 2β + 1)(β + 2α + 1) = αβ + 2α 2 + α + 2β 2 + 4αβ + 2β + β + 2α + 1 = 5αβ + 2(α 2 + β 2 ) + 3(α + β) + 1 7 37 3 = 5( − ) + 2( ) + 3( − ) + 1 2 4 2 5 =− 2
( x + 1)2 = 0 x = −1
`ä~ëëïçêâ=U=EéKNQF 1. x 2 − 3 x + 1 = 0 Sum of the roots = 3
2. α and β are the roots of x 2 + px + q = 0 . Sum of the roots = α + β = − p Product of the roots = αβ = q
Product of the roots = 1
( pα − q )2 + ( pβ − q )2 = p 2 α 2 − 2 pqα + q 2 + p 2β 2 − 2 pqβ + q 2
2. x 2 + 5 x − 3 = 0 Sum of the roots = −5
= p 2 (α 2 + β 2 ) − 2 pq(α + β) + 2 q 2
Product of the roots = −3
= p 2 [(α + β)2 − 2αβ] − 2 pq(α + β) + 2 q 2 = p 2 ( p 2 − 2 q ) − 2 pq( − p) + 2 q 2 = p 4 − 2 p2 q + 2 p2 q + 2q 2
3. 2 x 2 − 4 x − 7 = 0 Sum of the roots = 2
= p 4 + 2q 2
Product of the roots = −
7 2
`ä~ëëïçêâ=NM=EéKNUF 1. α and β are the roots of x 2 − x − 7 = 0 . Sum of the roots = α + β = 1
4. 5 x + (3k − 1) x + 2 k = 0 2
3k − 1 5 1 − 3k = 5
Sum of the roots = −
Product of the roots =
α β α 2 + β2 + = β α αβ
2k 5
Product of the roots = αβ = −7 1 1 β+α α+ +β+ = α+β+ α β αβ 1 = 1+ −7 6 = 7
37
Chapter 1 Quadratic Equations & Quadratic Functions
(b) For all real values of x,
1 1 )(β + ) α β 1 α β = αβ + + + β α αβ
(α +
1 ( x − )2 3 1 2 3( x − ) 3 1 2 4 3( x − ) − 3 3 1 2 4 3( x − ) − 3 3
1 α 2 + β2 = αβ + + αβ αβ (α + β)2 − 2αβ 1 + αβ αβ 1 − 2( −7) 1 = −7 + + −7 −7 65 =− 7 One required equation is 6 65 x2 − x − = 0 , i.e. 7 7 7 x 2 − 6 x − 65 = 0 = αβ +
2. Let α and α 2 be the two roots. Product of the roots = α(α 2 ) = 27
α 3 = 27 α =3
2 1 x− ) 3 3 2 2 1 2 2 = 3[ x − x + ( )2 − ( )2 − ] 3 6 6 3 1 2 4 = 3( x − ) − 3 3
≥−
4 3
4 3
(b) For all real values of x,
1 ( x − )2 ≥ 0 4 1 2 −4( x − ) ≤ 0 4 1 2 3 3 −4( x − ) − ≤ 0 − 4 4 4 3 −4 x 2 + 2 x − 1 ≤ − 4 3 4x2 − 2x + 1 ≥ 4
3 + 32 = − q q = −12
1. (a) 3 x 2 − 2 x − 1 = 3( x 2 −
≥0−
2. (a) −4 x 2 + 2 x − 1 1 1 = −4( x 2 − x + ) 2 4 1 1 1 2 1 = −4[ x − x + ( )2 − ( )2 + ] 2 4 4 4 1 2 3 = −4( x − ) − 4 4
Sum of the roots = α + α = − q
`ä~ëëïçêâ=NN=EéKOQF
≥0
4 ∴ The least value of 3 x 2 − 2 x − 1 is − . 3
2
3. Let α and 2α be the two roots. Product of the roots = α(2α ) = r Sum of the roots = α + 2α = −2q 2α 2 = r ............(1) 3α = −2 q .........(2) −2 q (2) : α = 3 −2 q Put α = into (1), 3 2q 2( − ) 2 = r 3 8q 2 =r 9 8q 2 = 9 r
≥0
`ä~ëëïçêâ=NO=EéKOUF y
1. (a)
−1
O
x 1
2
3
−1 −2 −3 −4
y = x 2 − 2x − 3
38
Chapter 1 Quadratic Equations & Quadratic Functions
y
(b)
y
(d)
y = x 2 + 2x + 4 4
1
x
O
1
2
3
4
5
3
6
−1
2
−2
1
−3
−2
−4
−1
x
O
1
2
−5
2. Since y = x 2 + (2 a + 1) x + a 2 intersects the x-axis at two distinct points, D > 0 .
y = −x 2 + 6x − 8
−6 −7
D = (2 a + 1)2 − 4 a 2 > 0
−8
4a 2 + 4a + 1 − 4a 2 > 0 a>−
1 4
y
(c)
3. Consider the discriminant of x 2 − 2 mx + m( m + 3) . O
x 1
2
3
4
5
6
D = ( −2 m)2 − 4 m( m + 3) = 4 m 2 − 4 m 2 − 12 m = −12 m
−1
m > 0 , −12 m < 0 Q D < 0 and the graph opens upwards
−2 −3
∴
−4 −5
`ä~ëëïçêâ=NP=EéKPOF
−6
1. (a)
−9
y = 6x − 9 − x
5 + 2x = 3 5 + 2x = 3 x = −1
−7 −8
x 2 − 2 mx + m( m + 3) is always positive for any real values of x if m > 0 .
2
(b)
or or
5 + 2 x = −3 x = −4
2 + 3x = 6 − 2 x 2 + 3 x = 6 − 2 x or 2 + 3 x = −(6 − 2 x ) or 2 + 3 x = −6 + 2 x 5x = 4 4 or x= x = −8 5
39
Chapter 1 Quadratic Equations & Quadratic Functions
2. (a)
`ä~ëëïçêâ=NR=EéKPQF
2x + 1 1 = 3− x 4
y
2x + 1 1 = 3− x 4 4(2 x + 1) = 3 − x 8x + 4 = 3 − x 1 x=− 9 (b)
2x + 1 1 =− 3− x 4 or 4(2 x + 1) = x − 3 or 8x + 4 = x − 3 or
x = −1
2 x + 1 = 9 + 3x
2
2 x + 1 = −9 − 3 x x = −2
2 x + 1 = 9 + 3 x or or x = −8 But 9 + 3 x ≥ 0 x = −2 ∴
1
`ä~ëëïçêâ=NQ=EéKPOF
−1
1. x 2 + 3 x − 4 = 6 x 2 + 3x − 4 = 6
or
x 2 + 3 x − 10 = 0 ( x + 5)( x − 2) = 0
or or
x 2 + 3 x − 4 = −6 x 2 + 3x + 2 = 0 ( x + 1)( x + 2) = 0
or x = −1, − 2 x = −5, 2 ∴ x = −5, −2, −1 or 2 2. 3( x + 1)2 − 7 x + 1 + 2 = 0 Let
a = x +1 ≥ 0 a = x +1 2
2
= ( x + 1)2 ∴ The equation becomes 3a 2 − 7a + 2 = 0 ( a − 2)(3a − 1) = 0 a = 2 or a =
Q
Q
x +1 = 2 x +1 = 2 x =1
1 3 1 x +1 = 3
y = 2 + 3x − 2x 2
3
1 3
or x + 1 = −2 or x = −3
x +1 =
1 3 2 4 or x=− x=− 3 3 2 4 ∴ x = −3, − , − or 1 3 3 or x + 1 = −
O
x 1
2