Question 1 :
Find all zeros of the polynomial x6 − 3x5 − 5x4 + 22x3 − 39x2 − 39x + 135, if it is known that 1 + 2i and √3 are two of its zeros.
Solution :
The four roots are 1 + 2i, 1 - 2i, √3 and -√3.
Quadratic equation whose roots are 1 + 2i and 1 - 2i.
Sum of roots = 2
Product of roots = 1 + 4 = 5
x2 - 2x + 5
Sum of roots = 0
Product of roots = -3
(x2 - 3) is a factor.
(x2 - 2x + 5) (x2 - 3) = x4 - 3x2 - 2x3 + 6x + 5x2 - 15
= x4 - 2x3 + 2x2 + 6x - 15
x2 - x - 9 = 0
x = (-b ± √b2 - 4ac)/2a
x = (1 ± √1 + 36)/2
x = (1 ± √37)/2
Hence the zeroes are 1 + 2i, 1 - 2i, √3, -√3, (1 ± √37)/2.
Question 2 :
Solve the cubic equation
(i) 2x3 − 9x2 +10x = 3
Solution :
2x3 − 9x2 +10x - 3 = 0
(x -1) is a factor
= 2x2 - 7x + 3
= 2x2 - 1x - 6x + 3
= x(2x - 1) - 3(2x - 1)
= (x - 3) (2x - 1)
(x - 1) (x - 3) (2x - 1) = 0
x - 1 = 0, x - 3 = 0, 2x - 1 = 0
x = 1, x = 3, x = 1/2
Hence the solutions are 1, 3 and 1/2.
(ii) 8x3 − 2x2 − 7x + 3 = 0.
Solution :
(x + 1) is a factor.
The other factors are 8x2 - 10x + 3
= 8x2 - 12x + 2x + 3
= 4x(2x - 3) -1(2x - 3)
= (4x - 1) (2x - 3)
By equating them to zero, we get
x + 1 = 0, 4x - 1 = 0, 2x - 3 = 0
x = -1, x = 1/4, x = 3/2
Hence the solutions are -1, 1/4 and 3/2.
Question 3 :
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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