41 Chapter 2
Review Exercise 2
Q6. Find value of k if roots of x 2- 3x + k + 1 =
0 differ by unity.
2 β=2 4 Sol: Comparing
x 2- 3x + k + 1 = 0 2 β=2with the quadratic equation
ax 2 + bx + c = 0 we have a = 1 , b = - 3, c =
k + 1
β
= 2 2
= 1 Sum of the roots α + β α + β = - ( - )3 1 = 3 1 =
3 =
- a bPutting
α + β = - a b2 + 1 =- 2 k Product of the roots α . β = c a Putting
3 =
- 2 k - k = 3 ×
2 α . β = k 1 + 1 = k +
1 ⇒ k = - 6According to the given condition Differ by unity
α - β =
1 x 3 + 6x 2 + 11x + 6 = 0 Using formula 4αβ = ( α + β ) 2 - ( α -
β )
2 putting the values
x 3 + 6x 2 + 11x + 6 = 0 4 ( k + 1 ) = ( 3 ) 2 -
( 1
) 2 4k + 4 = 9 -
1
Q ( x ) = x 2 + 3x + 2 R = 0 4k = 8 -
4 4k =4k = 4 4= 1 Q7 Find quadratic eq whose roots multiplicative
Sum Putting values of a,b,α & β
Q9. One root of cubic equation
is - 3. Use synthetic division to find the other roots. Sol: Since
-3 1 6 11 6 -3 -9 -6 1 3 2 0 Therefore and
To find the other roots take Q (x) = 0 x 2+ 3x + 2 =
0 x 2+ 2x + 1x + 2 =
0 x ( x + 2 ) + 1 ( x + 2 )
=
0 ( x + 1 )( x + 2 )
=
0
inverse of roots of
12x 2 - 17x + 6 = 0 Sol: Comparing
12x 2 - 17x + 6 = 0 with the quadratic equation
ax 2 + bx + c =
0 Either x + 1 =
0 x = -
1
or
x + 2 =
0 x = -
2 Hence the other roots are
- 1, - 2 we have a = 12 , b = - 17 , c =
6 α + β = - a b = - ( - 17 ) 12 =
17 12 α . β = c a =
12 6 Q10i). Solve x + y =
3
the system x 2 - 3xy + y 2
=
29
Sol: Since
x + y =
3 ( 1
)
x 2 - 3xy + y 2
=
29 ( 2 ) According to condition
Form equation (1) S = α 1 + β 1 =
β αβ+ α
Putting the values
17S = 12612
=
17 6 Putting the value of y in equation (2) we get Product of the roots P = α 1 . β 1 =
αβ 1 Putting the values P = 1 ÷ 12 6 = 12 6 = 2 The x 2- ( required Sumoftheroots equation ) x + is ( Productoftheroots given by
) = 0 x 2 - ⎛ │ ⎝ Q8. If one of the root of quadratic equation
Either Or
find the value of k
is 2, find the other root. Also
Substituting these values in equation (3)
Sol: Comparing with the quadratic equation we have 2 , k, 4
Solution Set Putting the values of a, c and α
17 6
⎞ │ ⎠ x + ( 2 ) = 0 × by 6 ⇒ 6x 2- 17x + 12 =
0
2x 2 + kx + 4 =
0 2x 2 + kx + 4 =0 ax 2 + bx + c = 0 a = b = c =
α . β =
c a
( x 2
- 3x 3 - x ) + ( 3 - x )
2 =
29
x 2 - 9x + 3x 2
+ ( 3 ) 2 - 2 ( 3 )( x ) + ( x )
2 =
29
x 2 - 9x + 3x 2 + 9 - 6x + x 2
=
29 x 2 + 3x 2 + x 2
- 9x - 6x + 9 - 29 =
0
5x 2
- 15x - 20 = 0 ÷
by 5 x 2- 3x - 4 =
0 x 2- 4x + 1x - 4 =
0 x ( x - 4 ) + 1 ( x - 4 )
=
0 ( x + 1 )( x - 4 )
=
0 x + 1 =
0 x = -
1
x - 4 =
0 x =
4
y 3 ( 1 ) y 3 1 y 4
= - -
y = 3 -
4 = +
y = -
1 =
= { ( - 1,4 ) , ( 4, -
1 ) }
x + y =
3 y = 3 -
x 3
( )
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