Let ax^{3} + bx^{2} + cx + d = 0 be any cubic equation and α, β and γ be the roots.
α + β + γ = -b/a
αβ + βγ + γ α = c/a
αβγ = -d/a
Example 1 :
Solve the following cubic equation whose roots are in arithmetic progression.
x^{3} - 12x^{2} + 39 x - 28 = 0
Solution :
When we solve cubic equation we will get three roots.
Since the roots are in arithmetic progression, the roots can be taken as given below.
p - q, p, p + q
Compare :
x^{3 }- 12x^{2 }+ 39x - 28 = 0 and ax^{3 }+ bx^{2 }+ cx + d = 0
a = 1, b = -12, c = 39, d = -28
Sum of the roots = -b/a
p - q + p + p + q = -(-12)/1
3p = 12
p = 4
4 is one of the roots. The other roots can be determined by solving the quadratic equation
x^{2 }- 8x + 7 = 0
x^{2 }- 8x + 7 = 0
(x - 1)(x - 7) = 0
x - 1 = 0 or x - 7 = 0
x = 1 or x = 7
Therefore the roots are 1, 4 and 7.
Example 2 :
Solve the following cubic equation whose roots are in geometric progression.
x^{3 }- 19x^{2} + 114x - 216 = 0
Solution :
When we solve cubic equation we will get three roots.
Since the roots are in geometric progression, the roots can be taken as given below.
p/q, p, pq
Compare :
x^{3 }- 19x^{2 }+ 114x - 216 = 0 and ax^{3 }+ bx^{2 }+ cx + d = 0
a = 1, b = -19, c = 114, d = -216
Product of roots = -d/a
p/q ⋅ p ⋅ pq = -(-216)/1
p^{3} = 216
p^{3} = 6^{3}
p = 6
6 is one of the roots. The other roots can be determined by solving the quadratic equation
x^{2 }- 13x + 36 = 0
(x - 4)(x - 9) = 0
x - 4 = 0 or x - 9 = 0
x = 4 or x = 9
Therefore the roots are 4, 6 and 9.
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