Chapter 2
19 Chapter 2
Exercise 2.1
Therefore, roots of the given equation are real {irrational} and unequal Exp2iii) Find nature of roots of 6 x 2 x- - 15 = 0 Quadratic Formula:
x =
- b ± b 2 2 a
- 4
ac Sol: Given 6 x 2 - x - 15 = 0 by comparing
a = 6, b = - 1, c = - 15 Nature of the roots of quadratic equation depends upon discriminant i.e., b 2 -
4 ac Discriminant = b 2 - 4 ac ( 1 ) 2
4 ( 6 )( 15 ) Roots are Real have 3 cases i). if b 2 - 4 ac = 0 i.e., 2 0 2 1 360
0
2
= - - - = +
x = - b ± a = - b ± a =
- a b roots are Real { rational } and equal
=361 =
19 2Therefore, roots of the given equation are real {rational} and unequal ii). if b 2 - 4 ac is perfect square Let b 2 - 4 ac =
d 2 Exp2iv) Find nature of roots of 4 x 2 + x + 1 =
0 Sol: Given 4 x 2 + x + 1 = 0 by comparing
i.e.,
x = - b ± 2 a d 2
=
- b 2
± a
d a = 4, b = 1, c = 1 Discriminant = b 2 - 4 ac roots are Real { Rational } and unequal iii). if b 2 - 4
ac is positive number e Let b 2 - 4 ac = e i.e., x b 2a
= ( 1 ) 2 -
4 ( 4 )( 1 ) = 1 -
16
=
- ± e = - 15 <
0 Therefore, roots of the given equation are complex conjugate {imaginary} and unequal roots are Real { irrational } and unequal iv) roots are Imaginary/ complex conjugate if b 2 - 4 ac is negative number i.e., -
f Exp3i Determine nature of roots of x 2 - 6 x + 9 =
0 And verify the result by solving them Sol: Given x 2 - 6 x + 9 = 0 by comparing Let b 2 - 4 ac = -
f a = 1, b = - 6, c = 9 i.e., x = - b ± - f 2 a =
- b ± i 2 a
f Discriminant = b 2 - 4 ac ( 6 ) 2 41 ( )( 9 ) roots are imaginary / complex conjugate
36 36 Exp 1. Find discriminant of x 2 + 9 x + 2 =
0 0 Sol: we have x 2 + 9 x + 2 = 0 by comparing a = 1, b = 9,c = 2 Discriminant = b 2 -
4 ac 9 241 ( )( 2 ) 81 8 73
= - - = - = Therefore, roots of the given equation are real {rational} and equal. Now verification x 2- 6 x
+ 9 =
0
= - = - = Exp2i). Find nature of roots of x 2 - 8 x + 16 = 0 Sol: Given x 2 - 8 x + 16 = 0 by comparing a = 1, b = - 8, c = 15 Discriminant = b 2 - 4 ac = ( - 8 ) 2 -
4 ( 1 )( 16
) x 2
- 3 x - 3 x
+ 9 =
0 x ( x - 3 ) - 3 ( x
- 3 )
=
0 ( x - 3 )( x
- 3 )
=
0 Either x- 3 = 0 or x- 3 =
0 x = 3 x = 3 so roots of given equation are rational & equal Exp3i Determine nature of roots of x 2 + 5 x + 6 = 0 And verify the result by solving them Sol: Given x 2 + 5 x + 6 = 0 by comparing = 64 -
64
a = 1, b = 5, c = 6 =
0 Therefore, roots of the given equation are real Discriminant = b 2 - 4 ac ( ) 2 ( )( ) {rational} and equal Exp2ii). Find nature of roots of x 2 + 9 x + 2 =
0 2 Sol: Given x 2 + 9 x + 2 = 0 by comparing a = 1, b = 9, c = 2 Discriminant = b 2 - 4 ac ( 9 ) 2 41 ( )( 2 ) 81 8 73 0
= - = - = = Therefore, roots of the given equation are real {rational} and unequal. Now verification
( ) ( ) ( )( )
5 41 6 25 24 1 1
x 2+ 5 x
+ 6 =
0 = -
x 2
+ 3 x + 2 x
+ 6 =
0 = -
x x + 3 + 2 x
+ 3 =
0 = >
x + 2 x
+ 3 =
0
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