Example 1 :
Solve the equation
x4+4x3+5x2+2x-2 = 0
which one root is -1 + i
Solution :
Let
f(x) = x4+4x3+5x2+2x-2
Since one of the root is complex number, the other root may be its conjugate.
So, α = -1 + i β = - 1 - i
By using these two roots we can find a quadratic equation which is the part of the original equation. First, let us find the quadratic equation.
General form of any quadratic equation:
x²-(α+β)x+αβ = 0
α = -1+i β = -1-i
Sum of roots (α+β) = -1+i-1-i
= - 2
Product of roots (αβ) = (-1+i) (-1-i)
= (-1)2 - i2
= 2
Therefore the required quadratic equation
x2-(-2)x+2 = 0
x²+2x+2 = 0
This is the part of the equation of polynomial of degree 4.
By solving this quadratic equation
x2+2x-1 = 0
we get other two roots.
a = 1, b = 2, c = -1
x = [- b ± √(b2-4 a c)]/2 a
x = [-2 ± √(22-4 (1)(-1))]/2 (1)
x = [- 2 ± √8]/2
x = [- 2 ± 2 √2]/2
x = -1±√2
Therefore the four roots are
-1 + i ,- 1 - i , - 1 + √2 , - 1 - √2.
Example 2 :
Solve the equation
6x4 − 35x3 + 62x2 − 35x + 6 = 0
Solution :
By suing synthetic division repeatedly, we may solve this problem.
2 and 3 are the solutions of the given polynomial. To know the other solutions, let us solve the quadratic equation.
6x2 - 5x + 1 = 0
6x2 - 2x - 3x + 1 = 0
2x (3x - 1) - 1(3x - 1) = 0
(2x - 1) (3x - 1) = 0
2x - 1 = 0 and 3x - 1 = 0
x = 1/2 and x = 1/3
Hence the solutions are 2, 3, 1/2 and 1/3
Example 3 :
Solve
x4 + 3x3 - 3x - 1 = 0
Solution :
1 and -1 are the solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation
x2 + 3x + 1 = 0
x = -b ± √(b2 - 4ac) / 2a
x = -3 ± √(9 - 4) / 2(1)
x = (-3 ± √5)/2
So, roots are -1, 1, (-3 ± √5)/2.
Example 4 :
Solve the equation 6x4 − 5x3 − 38x2 − 5x + 6 = 0 if it is known that 1/3 is a solution.
Solution :
1/3 and 3 are solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation.
6x2 + 15x + 6 = 0
6x2 + 12x + 3x + 6 = 0
6x(x + 2) + 3(x + 2) = 0
(6x + 3)(x + 2) = 0
x = -1/2 and x = -2
Hence the solutions are -1/2, -2, 3 and 1/3.
Example 5 :
Solve the equation x4 −14x2 + 45 = 0
Solution :
Let x2 = t
t2 − 14t + 45 = 0
t2 − 9t - 5t + 45 = 0
t(t - 9) - 5(t - 9) = 0
(t - 5) (t - 9) = 0
t = 5 and t = 9
x2 = 5, x2 = 9
x = ±√5, x = ±3
Hence the solutions are √5, -√5, 3 and -3.
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