- #1
anemone
Gold Member
MHB
POTW Director
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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.
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.
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.
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.