Question 1197055: one solution of z3+ A z^2 -z-14 =0 is z=-2-root 3 i where A is a real number. find A and other two solutions
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Theorem:
Let P(x) be a polynomial with real number coefficients.
If x = a+bi is a root of P(x), then so is x = a-bi.
This is known as the complex conjugate.
In this problem, the root z = -2-i*sqrt(3) pairs up with z = -2+i*sqrt(3) as its conjugate.
We'll use those two roots to find
z = -2-i*sqrt(3) or z = -2+i*sqrt(3)
z+2 = -i*sqrt(3) or z+2 = i*sqrt(3)
Squaring both sides of either equation leads to
(z+2)^2 = i^2*(sqrt(3))^2
(z+2)^2 = -1*3
(z+2)^2 = -3
Then we can get everything to one side and expand everything out
(z+2)^2 = -3
(z+2)^2+3 = 0
z^2+4z+4+3 = 0
z^2+4z+7 = 0
You can use the quadratic formula to verify that the roots of this equation are z = -2-i*sqrt(3) and z = -2+i*sqrt(3)
I'll leave this task to the student.
From here, we know that z^2+4z+7 is a factor of z^3+Az^2-z-14
which means
(z^2+4z+7)*Q(z) = z^3+Az^2-z-14
where Q(z) is some linear polynomial
We know it's linear because we need its leading term to multiply with z^2 to get z^3
You can probably see that the leading term of Q(z) must be z since z*z^2 = z^3
So we really have
Q(z) = z+B
where B is some real number
Now expand out the left hand side and group like terms
(z^2+4z+7)*Q(z) = z^3+Az^2-z-14
(z^2+4z+7)*(z+B) = z^3+Az^2-z-14
z(z^2+4z+7)+B(z^2+4z+7) = z^3+Az^2-z-14
z^3+4z^2+7z+Bz^2+4Bz+7B = z^3+Az^2-z-14
z^3 + (4z^2+Bz^2) + (7z+4Bz) + 7B = z^3+Az^2-z-14
z^3 + (4+B)z^2 + (7+4B)z + 7B = z^3+Az^2-z-14
Equating the coefficients gets us
4+B = A .... z^2 terms
7+4B = -1 ... z terms
7B = -14 .... constant
Solving either of the last two equations leads to B = -2
Then use this B value with 4+B = A to find A = 2
Therefore,
Q(z) = z+B = z-2
and
(z^2+4z+7)*Q(z) = z^3+Az^2-z-14
(z^2+4z+7)*(z+B) = z^3+2z^2-z-14
(z^2+4z+7)*(z-2) = z^3+2z^2-z-14
From there it's not too much of a leap to see that the factor z-2 gets us the real number root z = 2
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Answers:
A = 2
The three roots are:
z = -2-i*sqrt(3)
z = -2+i*sqrt(3)
z = 2
Answer by ikleyn(52816)  (Show Source):
You can put this solution on YOUR website! .
One solution of z3+ A*z^2 - z - 14 = 0 is z = -2-root(3)*i, where A is a real number.
Find A and other two solutions
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The solution in the post by math_tutor2020 may scare a reader - so complicated it is
with tons of calculations.
Meanwhile, the problem can be solved mentally in a very simple way, if to use Vieta's theorem.
Since the given equation is with real coefficients (given), its complex roots are conjugated complex numbers.
Since one such root is  =  (given), the other root is  =  .
Vieta's theorem says that the product of three roots of the given equation equals to the constant term
with the opposite sign. So
 = 14.
The product of the two complex conjugated numbers and is
 = 4 + 3 = 7.
Therefore, the third root of the given equation is  =  = 2.
Now the coefficient A equals to the sum of the three roots with the opposite sign, due to the same Vieta's theorem
A =  =  = -(-2-2+2) = 2.
ANSWER. A = 2. The other two roots are and 2.
Solved.
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Actually, the solution and the calculations are so simple that can be made mentally in the head.
The problems of this type are usually INTENDED, DESIGNED and EXPECTED to be solved in this way, using Vieta's theorem.
Then the solution is easy, elegant and provides fun for a student, showing a beauty of Math,
instead of making many tons of unnecessary complicated calculations.
And only in this context it has an educational value.
On Vieta's theorem see, for example, this Wikipedia article
https://en.wikipedia.org/wiki/Vieta%27s_formulas
This problem is a typical of a high school Math circle level.
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