26 Chapter 2
Exercise 2.2
=- 2a 2b
α + β
=
- a b( α + β ) 2 =
( 3a ) 2 α 2 + β 2 + 2 αβ = 9a 2 Putting the values 7 + 2 ( a 2 ) =
9
a
2
The Product of the roots Product = α . β = ⎛ │ │ ⎝ =
- b + ( - b ) 2 - ( b 2a b 22 -
- 4ac 4ac ) ⎞⎛ ││ ││ ⎠⎝ 2 ( 2a
) 2
- b - 2a b 2
- 4ac ⎞ │ │ ⎠
7 = 9 a 2 -
2
a
2
7 = 7a 2 a 2 = 1 a⇒ = ± 1 Exp13. Find value of k if roots of x 2 - 7 x + k = 0 =
+ b 2 - ( b 2- 4ac
) 4a
2
=b 2 - 4a.a
b 2
+ 4ac
=4a.a 4ac
differ by unity Sol: Given roots are differ by unit Assume that α , α + 1 be the roots of x 2 - 7 x + k = 0 ∴ S = α + α + 1 = - a b = - ( 1
- )7 = 7 =
ca
2 α+ 1 =
7 2 α= 7 -
1 Exp10i). Without solving find the sum and product of the roots of 2 x 2 - 3 x - 4 = 0 Sol: Given 2 x 2 - 3 x - 4 = 0 by comparing a = 2, b =- 3, c = - 4 ∴ sum of roots S = - a
b = - ( 2 - )3 =
3 2
2 α=6α =
3 ∴ P = α ( α + )1 = a c = k 1
= k putting α = 3 3 ( 3 + 1
) =
∴ Product of roots 2
4 2 k P = a
c = - = -
k
=
12 Exp13. If ,α β are roots of 9 x 2 - 27 x + k = 0 ,
Exp10ii). Without solving find the sum and product of the roots of 3 x 2 + 6 x - 2 = 0 Sol: Given 3 x 2 + 6 x - 2 = 0 by comparing a = 3, b = 6, c = -
2 Find the value of k such that 2 α + 5 β = 7 Sol: Given 9 x 2 - 27 x + k = 0 ∴ sum of roots S = - ab = - 3
6 = - 2 ∴ Product of roots P = a
c =
- 3 2Exp11. Find value of k so that sum of roots of
α + β
=
- ( - 9
27 ) α + β = 279
αβ = k 9.......(2)
α β+ = 3 .........(1) Given condition 2 α + 5 β =
7 2 α + 2 β + 3 β = 7 2 x 2 + kx + 6 = 0 is equal to three times product
2 ( α + β ) + 3 β = 7 putting of roots Sol: Given 2 x 2 + kx + 6 = 0 by comparing a = 2, b = k , c = 6 according to condition
2 ( 3 ) + 3 β
=
7 6 + 3 β
=
7
sum of roots = 3× Product of roots
3 β= 7 -
6 S =
3
P 3 β=1- a b = 3
a c
β
=
13 - b = 3 c putting the values - k=3 ( 6
) Put in eq (1) α + 1 3 = 3 k= -
18
α = 3 - 1 3 = 9 3 - 1 = 8 3 Exp12. Find value of a if sum of square of roots x 2 - 3 ax + a 2 = 0 is 7
eq (2) k 9= αβ Putting Sol: Given α 2 β+ 2 = 7 and x 2 - 3 ax + a 2 = 0 by comparing A = 1, B = - 3 a , C =
a 2 k9 = ⎛ │ ⎝ 8 3 1 3
∴
k
8 9 9
k8
⎞⎛ ││ ⎠⎝ ⎞ │ ⎠ P = αβ = a c = a 1 2
=
a 2 =∴ S = α + β = - a b = - ( - 1
3 a ) =
3 a ⇒ =
27 Exercise 2.3
Exp15. Find value of m and n if both sum and product of roots of mx 2 - 5 x + n = 0 Sol: Given mx 2 - 5 x + n = 0 by comparing a = m , b = - 5, c = n according to condition ∴ α + β = 10 ∴ αβ = 10 - ( m
- )5 =
10
5 =10 m10 5=m
m=
12Chapter 2
Q2. Find the value of k if sum of the roots of 2x 2 + kx + 6 =
0 is equal to product of roots. Sol: Comparing 2x 2 + kx + 6 =0 with the quadratic equation
ax 2 + bx + c = 0 we have a = 2 , b = k, c =
6 mn=10
According to the given condition Sum of the roots = Product of the roots i.e.,
n =
10
m
n = 10 ⎛ │ ⎝ m
.
1 2 ⎞ │ ⎠
∴ = 1 2
b c a a by a b c n
=
5
Substituting the values of b and c
Exercise 2.3 Q1i). Without solving equation, find the sum &
Q3. Find the value of k if the sum of square of product of roots of
the roots of is equal to 13. Sol: Comparing
Sol: Comparing with the quadratic equation
with the quadratic equation we have 4 , - 4, - 3
we have 1 , - 5k, 6k2 Sum of the roots Sum of the roots Putting values of a and b
Putting the values of a and b
Product of the roots
Product of the roots
Putting values
Putting the values of a and c
Q1ii). Without solving equation, find the sum
According to the given condition & product of roots of Sol: Comparing with the quadratic
equation we have 2 , 5, 6 Sum of the roots Putting the values of and we get
Putting the values of a and b
Product of the roots Putting values
Q4. For what value of k the roots of the
Q1iii). Without solving equation, find the sum & product of roots of Sol: Comparing
equation differ by unity. Sol: Comparing with the quadratic equation we have 1 , - 5, k with the quadratic equation Sum of the roots we have 3 , 2, - 5 Sum of the roots Putting the values of a and b
Putting the values of a and b
Product of the roots
Product of the roots Putting values
Putting the values of a and c
equation we have 2 , 5, 6 Sum of the roots Putting the values of and we get
Putting the values of a and b
Product of the roots Putting values
Q4. For what value of k the roots of the
Q1iii). Without solving equation, find the sum & product of roots of Sol: Comparing
equation differ by unity. Sol: Comparing with the quadratic equation we have 1 , - 5, k with the quadratic equation Sum of the roots we have 3 , 2, - 5 Sum of the roots Putting the values of a and b
Putting the values of a and b
Product of the roots
Product of the roots Putting values
Putting the values of a and c
α + β = - a bα + β = - ( )5 2 =
- 2 5
α . β = c a α . β = 2 6 =
3 3x 2 + 2x - 5 =
0 3x 2 + 2x - 5 =0 ax 2 + bx + c = 0 a = b = c =
α + β = - a bα + β = - ( )2 3 =
- 2 3 α . β = c a α . β = - 3 5 α 2 + β 2+ 2 αβ - 2 αβ =
13 α + β 2 - 2 αβ
= 13 ∴ a 2 + b 2
+ 2ab = a +
b 2 α + β .α β ( 5k ) 2 - 2 ( 6k 2
)
=
13
25k 2 - 12k 2
=
13 13k 2=
13 ⇒ k 2
=
1 ⇒ k = ±
1
x 2- 5x + k =
0 x 2- 5x + k =
0 ax 2 + bx + c = 0 a = b = c =
α + β α + β =
= - b- a ( - )5 1 = 5 1 = 5 α . β = c a α . β = k 1 =
k
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