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x=?x^3-x^2+8x+10=0 and x1=3i+1?
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You need to know that the complex solutions come in pairs, hence, `x = 1 - 3i` is also a solution to the given equation.
You may use the Vieta's formulas to find the missing solution `x_3` , such that:
`x_1 + x_2 + x_3 = -b/a => x_1 + x_2 + x_3 = -(-1)/1`
`x_1 + x_2 + x_3 = 1`
Replacing 1 + 3i for `x_1` and `1 - 3i` for `x_2` yields:
`1 + 3i + 1 - 3i + x_3 = 1`
Reducing the terms yields:
`2 + x_3 = 1 => x_3 = 1 - 2 => x_3 = -1`
Hence, evaluating the solutions to the given equation, using Vieta's formulas, yields `x_1 = 1 + 3i, x_2 = 1 - 3i, x_3 = -1.`
Posted by sciencesolve on March 20, 2013 at 6:31 PM (Answer #3)
Since the equation has a complex root, we'll recall the property of complex roots: If a complex number is the root of an equation, then it's conjugate is also the root of the equation.
So, we have as root the number z = a + bi => z' = a - bi is also the root of the equation.
Now, we'll verify if the complex numbers is the root of the equation by substituting it into the original equation.
We'll expand the cube using the formula:
(a+b)^3 = a^3 + b^3 + 3ab(a+b)
a = 1 and b = 3i
(1+3i)^3 = 1^3 + (3i)^3 + 3*1*3i*(1+3i)
(1+3i)^3 = 1 - 27i + 9i(1+3i)
We'll remove the brackets:
(1+3i)^3 = 1 - 27i + 9i - 27
We'll combine real parts and imaginary parts:
(1+3i)^3 = -26 + i(9-27)
(1+3i)^3 = -26 - 18i
We'll expand the square using the formula:
(a+b)^2 = a^2 + 2ab + b^2
(1+3i)^2 = 1^2 + 2*1*3i + (3i)^2
(1+3i)^2 = 1 + 6i - 9
(1+3i)^2 = -8 + 6i
We'll substitute the results into the expression (1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 = 0.
-26 - 18i - (-8 + 6i) + 8 + 24i + 10 = 0
We'll combine like terms:
(-26 + 8 + 8 + 10) + i(-18 - 6 + 24) = 0
0 + 0*i = 0
It is obvious that 1 + 3i is the root of the equation.
According to the rule, the conjugate of 1 + 3i, namely 1 - 3i, is also the root of the equation. So it is not necessary to verify if 1 - 3i is the root, since we've demonstrated that 1 + 3i is the root of the equation.
Posted by giorgiana1976 on May 30, 2011 at 7:09 PM (Answer #2)
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