28 Chapter 2
Exercise 2.3
According to the given condition Differ by unity
Q5. For what value of k the roots of the α - β =1 equation differ by three. Using formula 4αβ =
( α + β ) 2 - ( α -
β )
2 putting the values
Sol: Comparing with the quadratic equation 4k = ( 5 ) 2 -
( 1
)
2 we have 1 , - 9, k + 2
According to condition Let roots are α , α + 3 4k = 25 -
1 Sum of the roots 4k =24 k = 24 4= 6 Q4. For what value of k the roots of the
3 ( 1 )9 2 9 3
equation x 2- 5x + k =
0 differ by unity. Sol: Comparing
x 2- 5x + k = 0 with the quadratic equation we have 1 , - 5, k Let the roots are α , β α= + 1 Sum of the roots Putting the values of a, b &β
1 ( 1 )5 5 2 5 1
4 2 2 6 2 3 ax 2 + bx + c = 0 a = b = c =
α + β =- a bα . β =
c a x 2- 9x + k + 2 =
0 x 2- 9x + k + 2 =
0 ax 2 + bx + c = 0 a = b = c =
α + β = - a bα + β = - ( - )9 1 = 9 1 = 9 α . β = c a α . β = k + 1 x 2- 9x + k + 2 = 0 x 2- 9x + k + 2 =
0 ax 2 + bx + c = 0 a = b = c =α + α
+ = - - α= -
α
= =
Product of the roots α ( α + 3 ) = k + 1
2 putting the value of α
3 ( 3 + 3 ) = k
+
2 18 = k+
2 k= 18 -
2 k
= 16 Q6. If α,β be the roots of x 2- 5x + k =
0 , find k such that 3 α + 2 β = 12 Sol: Comparing
x 2- 5x + k = 0 with the quadratic equation
ax 2 + bx + c = 0 we have a = 1 , b = - 5, c =
k
So, α + α + β = 2 = k +
2 α - β = 3 4αβ = ( α + β ) 2 - ( α -
β ) 2 4 ( k + 2 ) = ( 9 ) 2 -
( 3
) 2 4k + 8 = 81 -
9 4k = 72 -
8 k = 64 4= 16
α + α
+ = - - =
α= -
α
= =
Product of the roots
Putting the values of a, c, ,α & β
α ( α + 1 )
=
k1 2 ( 2 + 1
)=
k k=
6
β - ( =1 - )5 - a b= Putting 5 1 =
5 the values of a and b
Now α . β =
c a Putting the values of a and c Q5. For what value of k the roots of the equation differ by three. Sol: Comparing
α . β = k 1 = k According to the given condition 3 α + 2 β
=
12 with the quadratic equation
α + 2 α + 2 β
=
12 we have 1 , - 9, k + 2 Sum of the roots α + 2 ( α + β
)
=
12 α + 2 ( 5 )
= 12 ∴ α + β
=
5 Putting the values of a and b
α+ 10 =
12 α= 12 -
10 α
=
2
Product of the roots
∴ α + β
=
5 2 + β = 5 ∴ α
=
2 Putting the values of a and c
β= 5 - 2 =
3 Now using
α . β = k According to given condition Roots are Differ by three Using formula
( 2 )( 3 ) k 2, 3 k 6 Q7. Find the value of m and n if both sum and product of the roots of the quadratic equation putting values
mx 2 - 3x - n = 0 are equal to .
Sol: Comparing mx 2 - 3x - n = 0 with the quadratic equation ax 2 + bx + c = 0 we have a =m ,b = - 3, c = - n Sum of the roots = 35
= ∴ α = β =
=
35
29 Exercise 2.3
α + β = - =
Chapter 2
35 b 3
then find value of symmetric function α 1 +
β 1 a 5 Sol: As ,α β are roots of ax 2 + bx + c = 0 Putting the values of a and b - ( - )3 m =
3 5 15 =
3m ⇒ m =
5 Product of the roots = 35 α β =
∴ α 1 + β 1 = β β × α 1 + β 1 ×
α α α 1 + β 1 = α αβ+ β put values
.35 c a =
3 5 Putting the values of a and c - m n =
3 5
= - a b ÷
a c
= - a b ×
a c
= - cbExp16vi): If ,α β are roots of ax 2 + bx + c = 0 - 5 n = 3 5
∴ m = 5 ⇒ n = -
3 Exp16i): If ,α β are roots of ax 2 + bx + c = 0 then find value of symmetric functionα β+ Sol: As ,α β are roots of ax 2 + bx + c =
0 then find value of symmetric function α 1 2 +
β 1 2 Sol: As ,α β are roots of ax 2 + bx + c = 0 ∴ α 1 2 + β 1 2 = β β α + β = - a b2 2 × α 1 2 + β 1 2 ×
α α
2
2 1 α 2 + β 1 2 = α α 2 2 + ββ 2 Exp16ii): If ,α β are roots of ax 2 + bx + c = 0 then find value of symmetric functionαβ
2
put values
=
α 2 + β 2
α + 2 2 β
αβ 2 - 2 αβ
Sol: As ,α β are roots of ax 2 + bx + c = 0 αβ =
a c= ( α + ( β αβ )
2
) - 2 2 αβ
putting values
Exp16iii): If ,α β are roots of ax 2 + bx + c =
0 = ⌈ │ │ ⌊ ⎛ │ ⎝ - ⎞ │ ⎠ 2 - ⌉ │ │ ⌋ then find value of symmetric function α 2 β+ 2 Sol: As ,α β are roots of ax 2 + bx + c = 0 α 2 + β 2 = α 2 + β 2 + 2 αβ -
2 αβ = ( α + β )2 - 2 αβ putting the values
2
222
2
2
2 2
2 2
2 2
2 2
2
2 ÷ ⎛ │ ⎝ ⎞ │ ⎠ = ⌈ │ ⌊ - × ⌉ │ ⌋ × = - ×
= - Formation of quadratic equation through roots Since ax 2 + bx + c = 0 ( )
b a 2
a c c a
b 2c a a a a a c
b 2ac a a c
= ⎛ │ ⎝ - b a ⎞ │ ⎠ - ⎛ │ ⎝ a c
⎞ │ ⎠
b 2ac
c
= b a - a c ×
a a
=
b - aac Exp16iv): If ,α β are roots of ax 2 + bx + c = 0 then find value of symmetric function α 3 β+ 3 Sol: As ,α β are roots of ax 2 + bx + c = 0 ∴ ( α + β ) 3 = α 3 + β 3 + 3 αβ ( α + β ) α 3 + β 3 = ( α + β ) 3 - 3 αβ ( α + β ) put values
3
33 2
3
3
2
22
x 2+ b a x + a
c =
0
x 2- ⎛ │ ⎝ - b a ⎞ │ ⎠
x + a
c = 0
x 2
- α + β x + αβ
=
0 x 2 - Sx + P = 0 Exp 17 Form a quadratic equation whose roots are 1 + 5 and 1 - 5 Sol:
Roots of quadratic eq are 1 + 5 &1 - 5 = ⎛ │ ⎝ Sum Product of roots of roots = S P = = 1 + ( 1 + 5 + 5 1 )( - 1 -
5 = 5 - a b ⎞ │ ⎠ - ⎛ │ ⎝ a c ⎞⎛ ││ ⎠⎝ - a
b
⎞ │ ⎠ = - a b + a bc ×
a a
2 ) 1 2( 5 )2 1 5 4
=
- b + a
3
33
abc
Exp16v): If ,α β are roots of ax 2 + bx + c =
0 = -
= - = - Since x 2 - Sx + P = 0 putting values 2( ) 2
PP
x - 2 x
+ - 4 =
0
x - 2 x
- 4 =
0
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