21 Chapter 2
Exercise 2.1
Q3i). For what value of k the roots of x 2+ kx + 9 = 0 are equal. Sol: Comparing x 2+ kx + 9 =
0 x = - 5 ± 2 with the quadratic equation ax 2 + bx + c = 0 we have a =1 ,b =k,c =9 As the roots are equal, therefore ∴ Discriminant = b 2- 4ac =
0 ( k ) 2
4 ( 1 )( 9 )
0
k 236 0 k 236 k 6
5 S S = ⎧ │ ⎨ │ ⎩ - 5 + 2 5 , - 5 - 2
5 ⎫ │ ⎬ │ ⎭ Unequal & irrational Q4ii). Determine whether 4x 2 + 12x + 9 = 0 have real roots and if so, find roots.
⇒ - =
Sol: Comparing 4x 2 + 12x + 9 = 0 ⇒ - = ⇒ = ⇒ = ± Q3ii). For what value of k the roots of 12x 2 + kx + 3 = 0 are equal.
with the quadratic equation ax 2 + bx + c = 0 we have a =4 ,b =12,c =9 To check the nature of the roots ∴ Discriminant = b 2-
4ac = ( 12 ) 2 -
4 ( 4 )( 9
) Sol: Comparing 12x 2 + kx + 3 =
0 = 144 -
144 with the quadratic equation ax 2 + bx + c =
0 =
0 we have a =12 ,b =k,c =3 As the roots are equal, therefore
As b 2- 4ac = 0 , therefor, the roots are real { rational } and Equal { Repeated } ∴ Discriminant = b 2- 4ac =
0 Since roots are real so we will solve further ⇒ ( k ) 2
- 4 ( 12 )( 3 )
=
0
⇒ k 2- 144 =
0 ⇒ k 2=
144 ⇒ k = ±
12
4x 2
+ 12x + 9 =
0 ( 2x ) 2 + 2 ( 2x )( 3 ) + ( 3 )
2
=
0 ( 2x + 3 )
2
=
0 ( 2x + 3 )( 2x + 3 )
=
0 Q3iii). For what value of k the roots of
2x + 3 =
0
2x + 3 =
0 x 2- 5x + k = 0 are equal.
Either
2x = -3or 2x = -3Sol: Comparing x 2- 5x + k = 0 with the quadratic equation ax 2 + bx + c =
0 x =
- 2
3 x =
- 2 3 we have a =1 ,b = - 5 ,c =k As the roots are equal, therefore ∴ Discriminant = b 2- 4ac =
0 Solution Set = ⎧ ⎨ ⎩ - 2 3⎫ ⎬ ⎭Repeated ⇒ ( - 5 ) 2 - 4 ( 1 )( k ) =
0 ⇒ 25 - 4k =
0 ⇒ 25 =
4k
⇒ k =
25 4 Q4i). Determine whether x 2+ 5x + 5 = 0 have real roots and if so, find roots. Sol: Comparing x 2+ 5x + 5 =
0 Q4iii). Determine whether 6x 2 + x - 2 = 0 have real roots and if so, find roots. Sol: Comparing 6x 2 + x - 2 = 0 with the quadratic equation ax 2 + bx + c = 0 we have a =6 ,b =1,c =- 2 To check the nature of the roots ∴ Discriminant = b 2-
4ac ( 1 ) 2
4 ( 6 )( 2
) with the quadratic equation ax 2 + bx + c =
0 1 48 we have a =1 ,b =5,c =5 To check the nature of the roots ∴ Discriminant = b 2-
4ac = - -
= + =49= 7 2= ( 5 ) 2 -
4 ( 1 )( 5
)
As b 2- 4ac is a perfect square, therefor, the roots are real “Unequal and rational” = 25 -
20
Since roots are real so we will solve further = 5 >
0 As b 2- 4ac > 0 , but not a perfect square, therefor, roots are real; Unequal and irrational Since the roots are real so we will solve further Using quadratic formula
( ) ( ) ( )( ) x b b 2 4ac 2a
6x 2+ x - 2 =
0
6x 2
+ 4x - 3x - 2 =
0 2x 3x + 2 - 1 3x + 2 =
0
2x - 1 3x + 2 =
0
= - ± - Putting values of a,b & c
2x - 1 =
0
3x + 2 =
0 Either
2x =1or 3x = -2x =
- ( 5 ) ± ( 5 ) 2 - 4 ( 1 )( 5 ) 2 ( 1
)
x =
2 1 x =
- 3 2 x =
- 5 ± 25 - 20 2
Solution Set = ⎧ ⎨ ⎩ 2 1 ,- 3 2 ⎫ ⎬ ⎭
Unequal and rational
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