MHB Solve for x⁴+16x-12=0 using effective methods

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Hi MHB,

This is a problem expected someone to solve it using the shortest route possible, and I've solved it, but using one most ordinary method that is very tedious and less gratifying.:( But I firmly believe this problem is designed and meant for practicing of the heuristic/problem solving skills, the thing is I just don't see there is any shortcut to tackle it.:mad:

If you happened to know there is another way to solve this problem, would you please show me? Thanks a bunch in advance.:)

Problem:

Solve $x^4+16x-12=0$.

My solution:

Descartes' Rule of Signs suggests there is at most one positive and one negative real roots, and since there is only one inflexion point (if we let $f(x)=x^4+16x-12=0$, then $f'(x)=4x^3+16$ which then implies $f'(x)=0$ iff $x=-4^{1/3}$) for the given function, from $f(1)=5,\,f(0)=-12,\,f(-3)=21$, Intermediate Value Theorem tells us the function $f(x)=x^4+16x-12=0$ has one positive real, one negative real and two complex roots.

If we first let the 4 roots be $a+bi,\,a-bi,\,c+\sqrt{d},\,c-\sqrt{d}$, Vieta's formula says the sum of the four roots is zero. Thus, $2a+2c=0,\,\rightarrow\,a=-c$.

The coefficient of $1$ from the given function allows us to further assume the four roots (2 of which real and 2 of which complex) as $1+bi,\,1-bi,\,-1-\sqrt{d},\,-1-\sqrt{d}$. Note that this can be wrong, as the $+1$ can stick to the real roots and the $-1$ be assigned to the complex roots. That said, we need to check our working to verify of our answer.

According to the Vieta's formula, it tells us again that

$(1+bi)(1-bi+(-1-\sqrt{d})+(-1-\sqrt{d}))+(1-bi)((-1-\sqrt{d})+(-1-\sqrt{d}))+(-1-\sqrt{d})(-1-\sqrt{d}))=0$

This simplifies to $(b^2-2d+1)+(bd-2b)i=0$, i.e.,

$d=\dfrac{b^2+1}{2}$ and substituting this into $bd-3b=0$ gives $b(b^2+1-6)=0$, or $b^2=5$ since $b\ne 0$ and hence $d=3$.

This generates the four roots to be $1+\sqrt{5}i,\,1-\sqrt{5}i,\,-1+\sqrt{3},\,-1-\sqrt{3}$.

I checked and realized this is a valid set of the answer.

But I am not satisfied by my method, I thought there ought to be another easy way to do this good problem. So, any feedback would be greatly appreciated.
 
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Try factoring it into two quadratics:

$$x^4+16x-12=\left(x^2+ax+b\right)\left(x^2+cx+d\right)$$
 
Oh My Goodness!

It actually can be rewritten as
$\begin{align*}x^4+16x-12&=x^4+4x^2+4-4(x^2-4x+4)\\&=(x^2+2)^2-(2(x-2))^2\\&=(x^2+2+2(x-2))(x^2+2-2(x-2))\\&=(x^2+2x+4)(x^2-2x+6)\\&=(x-(-1+\sqrt{3}))(x-(-1-\sqrt{3}))(x-(1+\sqrt{5}i))(x-(1+\sqrt{5}i))\end{align*}$

See how simple this solution that I found online is! :o

Edit: Hey MarkFL, you're right, I'm such a fool, I don't know where my head is...hehehe...
 
$x^4 + 16x - 12 = (x^4 + 4x^2 + 4 ) - (4x^2 - 16x + 16) = (x^2+2)^2 - (2(x-2) ))^2$. Since it is of form $a^2 - b^2$, we can rewrite that as $(a-b)(a+b)$, which yields a product of two quadratic as MarkFL indicated.

EDIT: You beat me to the post.
 
anemone said:
Oh My Goodness!

It actually can be rewritten as
$\begin{align*}x^4+16x-12&=x^4+4x^2+4-4(x^2-4x+4)\\&=(x^2+2)^2-(2(x-2))^2\\&=(x^2+2+2(x-2))(x^2+2-2(x-2))\\&=(x^2+2x+4)(x^2-2x+6)\\&=(x-(-1+\sqrt{3}))(x-(-1-\sqrt{3}))(x-(1+\sqrt{5}i))(x-(1+\sqrt{5}i))\end{align*}$

See how simple this solution that I found online is! :o

Edit: Hey MarkFL, you're right, I'm such a fool, I don't know where my head is...hehehe...

I did not see that it could be rewritten as the difference of two squares...yet. :D
 
anemone said:
Oh My Goodness!

It actually can be rewritten as
$\begin{align*}x^4+16x-12&=x^4+4x^2+4-4(x^2-4x+4)\\&=(x^2+2)^2-(2(x-2))^2\\&=(x^2+2+2(x-2))(x^2+2-2(x-2))\\&=(x^2+2x+4)(x^2-2x+6)\\&=(x-(-1+\sqrt{3}))(x-(-1-\sqrt{3}))(x-(1+\sqrt{5}i))(x-(1+\sqrt{5}i))\end{align*}$

See how simple this solution that I found online is! :o

Edit: Hey MarkFL, you're right, I'm such a fool, I don't know where my head is...hehehe...

How does one even notice something like that? :confused:
Trial and error of regrouping and adding and subtracting terms?
 
Rido12 said:
How does one even notice something like that? :confused:
Trial and error of regrouping and adding and subtracting terms?

One will know by experience, Rido12! :o Or, since we know the given function could be factored as $(x^2+ax+b)(x^2+cx+d)$, we can still figure out the values of $a,\,b,\,c,\,d$ through a little computation by comparing the coefficients.:)
 
In case anyone wants a rigorous approach, the natural resolvents coming from a quartic are

$$\alpha_1 = (x_1 + x_2)(x_3 + x_4) \\ \alpha_2 = (x_1 + x_3)(x_2 + x_4) \\ \alpha_3 = (x_1 + x_4)(x_2 + x_3)$$

Where $x_i$ are the roots of the quartic. Note that these are left unchanged by the $4! = 24$ permutation of the roots. If we take the quartic in OP, it is not too hard to verify that $\alpha_i$s satisfies the cubic

$$x^3 + 48x + 256 = 0$$

This, however, factors as $(x + 4)(x^2 - 4x + 64) = 0$ by rational root theorem (indeed, this indicates that the Galois group of this polynomial is $D_4$ and shouldn't involve too much radicals). The roots are thus

$$\alpha_1 = -4 \\ \alpha_2 = 2(1 - i\sqrt{15}) \\ \alpha_2 = 2(1 + i\sqrt{15})$$

By some tedious computations, one retrieves the original roots of the quartic back

$$2x_1 = \sqrt{-\alpha_1} + \sqrt{-\alpha_2} + \sqrt{-\alpha_3} \\ 2x_2 = \sqrt{-\alpha_1} - \sqrt{-\alpha_2} - \sqrt{-\alpha_3} \\ 2x_3 = -\sqrt{-\alpha_1} + \sqrt{-\alpha_2} - \sqrt{-\alpha_3} \\ 2x_4 = -\sqrt{-\alpha_1} - \sqrt{-\alpha_2} + \sqrt{-\alpha_3}$$

Which are, after the application of the identities $\sqrt{-2\left (1-i\sqrt{15}\right)} + \sqrt{-2\left(1+i\sqrt{15}\right)} = 2 \sqrt{3}$ and $\sqrt{-2\left(1-i\sqrt{15}\right)} - \sqrt{-2\left(1+i\sqrt{15}\right)} = i2 \sqrt{5}$ and appropriate signs (this is the real pain in the neck) are

$$x_1 = -1 + \sqrt{3} \\ x_2 = -1 - \sqrt{3} \\ x_3 = 1 + i\sqrt{5} \\ x_4 = 1 - i\sqrt{5}$$

These methods are much more practical while doing calculations of these sort, although they do not describe describe any of the beauties of manipulation whatsoever.
 
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