For the quadratic equation which is in the form,
ax2 + bx + c = 0
if the quadratic equation has real and equal roots, it must satisfy the condition
b2 - 4ac = 0
Find the values of k for which the given quadratic equations have real and equal roots.
Problem 1 :
4x2 + kx + 9 = 0
Solution :
Since the roots are real and equal, b2 - 4ac = 0
From the equation given above,
a = 4, b = k and c = 9
k2 - 4(4)(9) = 0
k2 - 144 = 0
k2 = 144
k = √144
k = ±12
Then the possible values of k are -12 and 12.
Problem 2 :
kx2 - 5x + k = 0
Solution :
Since the roots are real and equal, b2 - 4ac = 0
From the equation given above,
a = k, b = -5 and c = k
(-5)2 - 4(k)(k) = 0
25 - 4k2 = 0
4k2 = 25
k2 = 25/4
k = √(25/4)
k = ±5/2
Then, the values of k are -5/2 and 5/2.
Problem 3 :
x2 - 7(3+k) + 4 - 2x(1 + k) = 0
Solution :
x2 - 7(3+k) + 4 - 2x(1 + k) = 0
The given quadratic equation is not in the standard form, by arranging it.
x2 - 2x(1 + k) - 7(3+k) + 4 = 0
From this equation, we get
a = 1, b = -2(1 + k) and c = - 7(3+k) + 4
c = -21 - 7k + 4
c = -17 - 7k
[-2(1 + k)]2 - 4(1)(-17 - 7k) = 0
4(1 + k)2 + 68 + 28k = 0
Using algebraic identity, finding the expansion
4(1 + 2k + k2) + 68 + 28k = 0
4 + 8k + 4k2 + 68 + 28k = 0
4k2 + 28k + 8k + 68 + 4 = 0
4k2 + 36k + 72 = 0
Dividing the entire equation by 4, we get
k2 + 9k + 18 = 0
Using the method of factoring, we can solve the equation above.
k2 + 3k + 6k + 18 = 0
k(k + 3) + 6(k + 3) = 0
(k + 3) (k + 6) =0
Equating each factor to 0, we get
k = -3 and k = -6
Problem 4 :
Consider x2 - 2x + m = 0. Find the discriminant, hence find the values of m for which the equation has
a) repeated roots
b) 2 distinct real roots
c) no real roots
Solution :
x2 - 2x + m = 0
a = 1, b = -2 and c = m
a) roots are repeated = roots are equal
Then b2 - 4ac = 0
The given quadratic equation is not in the standard form, by arranging it.
(-2)2 - 4(1)(m) = 0
4 - 4m = 0
4m = 4
m = 1
When m = 1, the quadratic equation will have repeated roots.
b) If the quadratic equation has two distinct real roots, then
b2 - 4ac > 0
(-2)2 - 4(1)(m) > 0
4 - 4m > 0
-4m > -4
Dividing by -4 on both sides, we get
m < 1
c) If the quadratic equation has no real roots, then
b2 - 4ac < 0
(-2)2 - 4(1)(m) < 0
4 - 4m < 0
-4m < -4
Dividing by -4 on both sides, we get
m > 1
Problem 5 :
If -4 is a root of the quadratic equation x2 + px - 4 = 0 and the quadratic equation 12x2 + 4px + k = 0 has equal roots, the find the value of k.
Solution :
For the quadratic equation, x2 + px - 4 = 0 we have a root which is -4.
Let α and β be roots of the quadratic equation. Then α = -4
Sum of roots (α + β) = -b/a
Product of roots α β = c/a
a = 1, b = p and c = -4
α + β = -p/1 ==> -p
α β = -4/1 ==> -4
-4 + β = -p-----(1)
-4β = -4
β = 1
Applying the α = -4 and β = 1 in (1), we get
-4 + 1 = -p
p = 3
Applying the value of p in the equation 12x2 + 4px + k = 0
12x2 + 4(3)x + k = 0
12x2 + 12x + k = 0
Since the roots of this quadratic equation are equal,
b2 - 4ac = 0
a = 12, b = 12 and c = k
(12)2 - 4(12)(k) = 0
144 - 48k = 0
-48k = -400
k = 400/48
k = 25/3
Problem 6 :
If the equation (1+k2)x2 + 2kqx + (q2 - p2) = 0 has equal roots, then shown that q2 = p2(1 + k2).
Solution :
(1+k2)x2 + 2kqx + (q2 - p2) = 0
a = 1+k2, b = 2kq and c = q2 - p2
Since the roots of this quadratic equation are equal,
b2 - 4ac = 0
(2kq)2 - 4(1+k2) (q2 - p2) = 0
4k2q2 - 4(q2 - p2 + k2q2 - k2p2) = 0
4k2q2 - 4q2 + 4p2 - 4k2q2 + 4k2p2 = 0
4p2 + 4k2p2 = 4q2
Dividing by 4, we get
p2 + k2p2 = q2
Factoring p2, we get
p2 (1 + k2) = q2
Hence proved
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM