Problem 1 :
If the sum of the roots of
4x^{2} + kx – 7 = 0 is 3,
Find the value of k.
Solution :
4x^{2} + kx – 7 = 0
Sum of roots (α + β) = -b/a
α + β = 3
-k/4 = 3
-k = 12
k = -12
Problem 2 :
Find the value of K if the sum of the roots of equation
(2k – 1)x^{2} + (4k – 1) x + (K + 3) = 0 is 5/2
Solution :
(2k – 1)x^{2} + (4k – 1) x + (K + 3) = 0
a = 2k – 1, b = 4k – 1 and c = K + 3
Sum of roots = -b/a
-(4k - 1)/(2k - 1) = 5/2
-2(4k - 1) = 5(2k - 1)
-8k + 2 = 10k - 5
-8k - 10k = -5 - 2
-18k = - 7
k = 7/18
Problem 3 :
For what value of K the roots of the following equations are equal :
(i) kx^{2} + 4x + 3 = 0, (ii) 2x^{2} + 5x + k = 0
Solution :
(i) kx^{2} + 4x + 3 = 0
a = k, b = 4 and c = 3
Sum of roots = -4/k
Product of roots = 3/k
Here α = β
α + α = -4/k 2α = -4/k α = -4/2k α = -2/k |
α ⋅ α = 3/k α^{2} = 3/k. (-2/k)^{2} = 3/k 4/k^{2} = 3/k 4/k = 3 k = 4/3 |
(ii) 2x^{2} + 5x + k = 0
Solution :
a = 2, b = 5 and c = k
Sum of roots = -5/2
Product of roots = k/2
Here α = β
α + α = -5/2 2α = -5/2 α = -5/4 |
α ⋅ α = k/2 α^{2} = k/2 (-5/4)^{2} = k/2 25/16 = k/2 k = (25/16) x 2 k = 25/8 |
Problem 4 :
If the difference of the roots of 6x^{2} – 23x + c = 0 is 5/6, find the value of c
Solution :
Difference of roots = α - β = √(α + β)^{2} - 4αβ
α - β = 5/6
α + β = -(-23)/6 ==> 23/6
α β = c/6
α - β = √(23/6)^{2} - 4(c/6)
(5/6)^{2} = 529/36 - (2c/3)
(2c/3) = (529/36) - (25/36)
(2c/3) = (529 - 25)/36
c = (504/36) x (3/2)
c = 21
Problem 5 :
If the difference of the roots of x^{2} – 7x + k – 4 = 0 is 5, find the value of k and the roots.
Solution :
x^{2} – 7x + k – 4 = 0 ----(1)
α - β = 5
α + β = -(-7)/1 ==> -7
α β = (k -4)/1 ==> k - 4
α - β = √(α + β)^{2} - 4αβ
5 = √(-7)^{2} - 4(k - 4)
5^{2} = 49 - 4k + 16
25 = 65 - 4k
4k = 65 - 25
4k = 40
k = 10
By applying the value of k in (1), we get
x^{2} – 7x + 10 – 4 = 0
x^{2} – 7x + 6 = 0
(x - 1)(x - 6) = 0
x = 1 and x = 6
So, the roots are 1 and 6.
Problem 6 :
Find the value of k given that if one root of
9x^{2} – 15x + k = 0
exceeds the other by 3. Also find the roots.
Solution :
9x^{2} – 15x + k = 0
α = β + 3
α + β = -(-15)/9 ==> 5/3
α + β = 5/3
α β = k/9
Applying the value of α in (1), we get
β + 3 + β = 5/3
2β + 3 = 5/3
2β = (5/3) - 3
2β = -4/3
β = -2/3
α = (-2/3) + 3
α = 7/3
So, the roots are -2/3 and 7/3.
Problem 7 :
The product of the roots of the equation
(k + 1)x^{2} + (4k + 3)x + (k – 1) = 0
is 7/2
Solution :
(k + 1)x^{2} + (4k + 3)x + (k – 1) = 0
a = k + 1, b = 4k + 3 and c = k - 1
Product of roots = c/a
(k - 1)/(k + 1) = 7/2
2(k - 1) = 7(k + 1)
2k - 2 = 7k + 7
2k - 7k = 7 + 2
-5k = 9
k = -9/5
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