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Solve the equation

$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$

Having a complex root of modulus $1$.

To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.

Jean-Claude Arbaut
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user146270
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3 Answers3

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HINT:

$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = (x^2 + 2)(3x^2 + 2x + 3)$$

barak manos
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  • Is there anything other than observation involved here? – lab bhattacharjee Aug 29 '14 at 13:29
  • @labbhattacharjee this is a useful hint. If the product is equal to $0$ then we have $x^2 + 2 = 0$ and $3x^2 + 2x + 3 = 0$. By applying he quadratic formula to the system of equations, we obtain the values of $x$, namely all the four roots of the general equation. – Mr Pie Jan 31 '18 at 09:23
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Note that you have an equation with integer coefficients (real coefficients would do), so that any complex root $a$ of modulus $1$ implies a conjugate root $\bar {a }$ with $a\bar a=1$. You will therefore have a real factor of the form $x^2-bx+1$ with $b=a+\bar a$. Making obvious deductions from the leading and constant terms, there is a real factorisation of the original quartic of the form $$3x^4 + 2x^3 + 9x^2 + 4x + 6 = (x^2-bx+1)(3x^2+cx+6)$$

Equating coefficients of $x^3$ and $x$ gives $c-3b=2$ and $c-6b=4$, so that $c=0, b=-\frac 23$.

That gives two quadratic equations to be solved. [Note the factor $3$ can be redistributed to give a factorisation with integer coefficients]

Mark Bennet
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Let the root be $\cos y+i\sin y,$

Using Complex conjugate root theorem, $\cos y-i\sin y$ must be another root

So, if the four roots are $\cos y\pm i\sin y,u, v$

using Vieta's formula, $(\cos y+i\sin y)(\cos y-i\sin y)u\cdot v=\dfrac63\implies v=\dfrac2u$

So we have $$3[x-(\cos y+i\sin y)][x-(\cos y-i\sin y)](x-u)\left(x-\dfrac2u\right)=3x^4 + 2x^3 + 9x^2 + 4x + 6$$

$$\iff3[(x^2-2x\cos y+1)]\left[x^2-\left(u+\frac2u\right)+2\right]=3x^4 + 2x^3 + 9x^2 + 4x + 6$$

$$\iff3\left[x^4-x^3\left(2\cos y+u+\frac2u\right)+x^2\left[1+2+2\cos y\left(u+\frac2u\right)\right]+\cdots\right]=3x^4 + 2x^3 + 9x^2 + 4x + 6$$

Equating the coefficients of $x^3,x^2$

$$2\cos y+u+\frac2u=-\frac23\ \ \ \ \ (1)$$ and $$1+2+2\cos y\left(u+\frac2u\right)=\frac93\iff2\cos y\left(u+\frac2u\right)=0$$

If $\cos y=0\implies\sin y=\pm1\implies \cos y\pm\sin y=\pm i$ which does not satisfy the given equation

So, $u+\dfrac2u$ must be $0$

and from $(1),2\cos y=-\dfrac23\iff\cos y=-\dfrac13\implies\sin y=\pm\dfrac{2\sqrt2}3$

Observation: $x^2+2$ is a factor of $$3x^4 + 2x^3 + 9x^2 + 4x + 6$$

lab bhattacharjee
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