Determine the expression of the equation 2x^3-x^2+ax+b=0 if a solution is 1+i .
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We have the equation: 2x^3 - x^2 + ax + b = 0 with one of its solutions given as 1 + i.
Substituting x with 1 + i,
2*(1 + i)^3 - (1 + i)^2 + a(1 + i) + b = 0
=> 2*( 1 + 3i^2 + 3i + i^3) - 1 - i^2 - 2i + a + ai + b = 0
=> 2*( 1 - 3 + 3i - i) - 1 + 1 - 2i + a + ai + b = 0
=> 2*( - 2 + 2i ) - 2i + a + ai + b = 0
=> -4 + 4i - 2i + a + ai + b = 0
=> -4 + a + b + i( 2 + a) = 0
Equating the real and complex coefficients, we have
-4 + a + b = 0
2 + a = 0
=> a = -2
substituting a = 6 in -4 + a + b = 0
=> -4 - 2 + b = 0
=> b = 6
The required equation is 2x^3 - x^2 - 2x + 6 = 0
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In other words, we'll have to determine a and b.
The equation has 3 solutions and 2 of them are complex, the 3rd solution being a real one.
If the equation has has a complex solution, that means that the conjugate of the complex solution is also a solution for equation.
So, the given equation has as solutions:
x1 = 1 + i and x2 = 1 - i
The polynomial 2x^3-x^2+ax+b is divided by (x - 1 - i)(x -1 + i).
P(1 + i) = 2(1+i)^3 - (1+i)^2 + a(1+i) + b = 0
(1 + i)^2 = 1 + 2i - 1 = 2i
(1+i)^3 = 2i(1+i) = -2 + 2i
P(1 + i) = -4 + 4i - 2i + a + ai + b = 0
-4 + a + b + i(a + 2) = 0
-4 + a + b = 0
a + b = 4
a + 2 = 0
a = -2
b = 4 - a
b = 4 + 2
b = 6
The expression of the equation is: 2x^3 - x^2 - 2x + 6 = 0.
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