If α, β, are the roots of the quadratic equation, then the form of the quadratic equation as
x2 – (α + β)x + αβ = 0
Where,
α + β = sum of roots
αβ = product of roots
Problem 1 :
Find k if the difference between the roots of the quadratic equation
x2 – 4x + k = 0 is 2.
Solution :
x2 – 4x + k = 0
a = 1, b = -4, c = k
α – β = 2 ----- (1)
α + β = -b/a α + β = -(-4)/1 α + β = 4 ----- (2) |
α – β = 2 α + β = 4 (1) + (2) 2α = 6 α = 6/2 α = 3 |
Products of the roots = αβ = c/a
3 ⋅ β = k/1
β = k/3
α – β = 2
3 – (k/3) = 2
-k/3 = 2 – 3
-k/3 = -1
k = 3
So, the value of k is 3.
Problem 2 :
Find the value of p such that the difference of the roots of the equation
x2 – px + 8 = 0 is 2.
Solution :
x2 – px + 8 = 0
a = 1, b = -p, c = 8
Difference between the roots = α – β = 2 ----- (1)
Sum of the roots = α + β = -b/a
α + β = -(-p)/1
α + β = p ----- (2)
Product of roots = α β = 8
(1) + (2)
2α = p + 2
α = (p + 2)/2
Applying the value of α = (p + 2)/2 in α β = 8
β = 8 ⋅ 2/(p + 2)
β = 16/(p + 2)
Applying the values of α and β in (1), we get
(p + 2)/2 - 16/(p + 2) = 2
So, the value of p is -6 or 6.
Problem 3 :
Find the value of k such that the difference of the roots of the equation
2kx2 – 20x + 21 = 0 is 2.
Solution :
2kx2 – 20x + 21 = 0
a = 2k, b = -20, c = 21
Difference between the roots = α – β = 2 ----- (1)
α = 2 + β
α + β = -b/a α + β = -(-20)/2k α + β = 10/k ----- (2) |
α β = c/a α β = 21/2k ---(3) |
Applying the value of α in (2) and (3), we get
2 + β + β = 10/k 2 + 2β = 10/k 2β = (10/k) - 2 β = (5/k) - 1 |
(2 + β) β = 21/2k Applying β [2 + (5/k) - 1][(5/k) - 1] = 21/2k ((5/k) - 1)((5/k) + 1) = 21/2k 25/k2 - 1 = 21/2k |
Problem 4 :
Find k so that one root of the equation 2x2 – 16x + k = 0 is twice the other. (Hint : One root = α, Other root = 2α)
Solution :
Given, 2x2 – 16x + k = 0
a = 2, b = -16, c = k
α = α, β = 2α
α + β = -b/a α + 2α = -(-16)/2 = 16/2 3α = 8 α = 8/3 |
αβ = c/a α(2α) = k/2 2α2 = k/2 α2 = k/4 |
By applying the value of α, we get
(8/3)2 = k/4
64/9 = k/4
k = (64 ⋅ 4)/9
k = 256/9
So, the value of k is 256/9.
Problem 5 :
Find k so that one root of the equation
k(x – 1)2 = 5x – 7
is twice the other.
Solution :
k(x – 1)2 = 5x – 7
k(x2 + 1 – 2x) – 5x + 7 = 0
kx2 + k – 2xk – 5x + 7= 0
x2(k) - x(2k + 5) + k + 7 = 0
a = k, b = 2k + 5, c = k + 7
α = α, β = 2α
α + β = -b/a α + 2α =( 2k + 5)/k 3α = (2k + 5)/k α = (2k + 5)/3k ---(1) |
αβ = c/a α(2α) = (k + 7)/k 2α2 = (k + 7)/k α2 = (k + 7)/2k ---(2) |
By applying α in (2), we get
Equating each factor to zero, we get
k = -25 and k = 2
Problem 6 :
Find k so that the roots of the quadratic equation
2x2 + 3x + k = 0
are equal.
Solution :
Given, 2x2 + 3x + k = 0
a = 2, b = 3, c= k
α = α, β = α
α + β = -b/a α + α = -3/2 2α = -3/2 α = -3/2 × 1/2 α = -3/4 |
αβ = c/a α × α = k/2 -3/4 × -3/4 = k/2 9/16 = k/2 9/16 × 2 = k 9/8 = k |
So, the value of k is 9/8.
Problem 7 :
If 1 – i and 1 + i are the roots of the equation
x2 + ax + b = 0
where a, b ∈ r, then find the values of a and b.
Solution :
Given, x2 + ax + b = 0
a = 1, b = a, c = b
α = 1 - i, β = 1 + i
α + β = -b/a 1 – i + 1 + i = -a 2 = -a a = -2 |
αβ = c/a (1 – i) × (1 + i) = b 1 – i2 = b 1 + 1 = b 2 = b |
So, the
values of a and b is -2 and 2 respectively.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM