20 Chapter 2
Exercise 2.1
Either x+ 2 = 0 or x+ 3 =
0 ax 2 + bx + c = 0 we get a =1 ,b = - 4,c =13 x = - 2 x = - 3 so roots of given equation are rational & unequal Exp4 without solving determine nature of roots
∴ Discriminant = b 2-
4ac = ( - 4 ) 2 -
4 ( 1 )( 13
) of 3 x 2 - 4 x + 6 =
0 = 16 -
52 Sol: Given 3 x 2 - 4 x + 6 = 0 by comparing
= -
36 a = 3, b = - 4, c =
6 Q1ii). Find discriminant of 4x 2 - 5x + 1 = 0 Discriminant = b 2 - 4 ac Sol: Since 4x 2 - 5x + 1 = 0 = ( - 4 ) 2 -
4 ( 3 )( 6
) Comparing with the quadratic equation = 16 -
72
ax 2 + bx + c = 0 we have a = 4,b = - 5,c =1
= - 56 <
0 Therefore, roots of the given equation are complex conjugate or imaginary and unequal Exp5 without solving determine nature of roots of 2 x 2 - 7 x = -
1 ∴ Discriminant = b 2-
4ac = ( - 5 ) 2 -
4 ( 4 )( 1
) = 25 -
16 =
9
Sol: Given 2 x 2 - 7 x = -
1 Q1iii). Find discriminant of x 2+ x + 1 = 0 Or 2 x 2 - 7 x + 1 = 0 by comparing a = 2, b = - 7, c = 1 Discriminant = b 2 - 4 ac ( 7 ) 2 4 ( 2 )( 1 ) 49 8 41 0
Sol: Since x 2+ x + 1 = 0 Comparing with the quadratic equation ax 2 + bx + c = 0 we have a =1 ,b = 1,c =1
= - - = - = > Therefore, roots of the given equation are real ∴ Discriminant = b 2-
4ac = ( 1 ) 2 -
4 ( 1 )( 1
) = 1 -
4 = -
3
{Irrational} and unequal
Q2i). Examine nature of roots of 3x 2 - 5x + 1 = 0 Exp6i). Determine the value of k for which
Sol: Since 3x 2 - 5x + 1 = 0 given kx 2 + 4 x + 1 = 0 have real roots. Sol: we have kx 2 + 4 x + 1 = 0 by comparing a = k , b = 4, c =
1 Comparing with the quadratic equation
ax 2 + bx + c = 0 we have a =3 ,b = - 5,c =1 ∴ Discriminant = b 2- 4ac Discriminant = b 2 - 4 ac ≥
0 = ( - 5 ) 2 -
4 ( 3 )( 1
) ( 4 ) 2 4 ( )( 1 ) 0
= 25 -
12 16 4 0
= 13 >
0
16 4 4
- k
≥ - k≥ ≥k ≥ kOr k ≤ 4 Exp6ii). Determine the value of k for which given 2 x 2 + kx + 3 = 0 have real roots. Sol: we have 2 x 2 + kx + 3 = 0 by comparing a = 2, b = k , c = 3 Discriminant = b 2 - 4 ac ≥
0 As b 2- 4ac > 0 , but not a perfect square,
therefor, roots are real {irrational} & Unequal Q2ii). Examine nature of roots of 6x 2 + x - 2 = 0 Sol: Since 6x 2 + x - 2 = 0 Comparing with the quadratic equation ax 2 + bx + c = 0 we have a = 6,b = 1,c =-2 ∴ Discriminant = b 2-
4ac = ( 1 ) 2
- 4 ( 6 )( -
2
) ( k) 2
- 4 ( 2 )( 3 ) ≥
0
= 1 +
48 =49k2- 24 ≥
0
=
7 2k
2
≥
24
As b 2- 4ac is a perfect square,
k2
≥
24
k
≥
2 6
Therefor, roots are real;{rational} & Unequal Q2iii)Examine nature of roots of 3x 2 + 2x + 1 = 0 Sol: Since 3x 2 + 2x + 1 = 0 k± ≥
2 6 Comparing with the quadratic equation Either k ≥ 2 6 or k- ≥
2 6 k ≤ - 2 6 Exercise 2.1
ax 2 + bx + c = 0 we have a =3 ,b = 2,c =1 ∴ Discriminant = b 2-
4ac = ( 2 ) 2 -
4 ( 3 )( 1
) = 4 -
12 Q1i). Find discriminant of x 2- 4x + 13 =
0 = - 8 <
0 Sol: Since x 2- 4x + 13 =
0 As b 2- 4ac < 0 , Therefor, roots are Comparing with the quadratic equation
imaginary & unequal or Complex conjugate
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