# Problem Of The Week #369 Jun 4th, 2019

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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If the equation $ax^2+(c-b)x+(e-d)=0$ has real roots greater than 1, show that the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real root.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solution!

1. castor28
2. Olinguito

Solution from castor28 :
As dividing the equations by $a$ does not change their roots, we may assume that $a=1$ (the use of the plural in “real roots” implies that $a\ne0$). The equations become:
\begin{align*}
f(x)&=x^2 + (c-b)x + (d-e) = 0\\
g(x)&= x^4 + bx^3 +cx^2 + dx + e = 0
\end{align*}
As both roots of $f(x)$ are greater than $1$, we have:
\begin{align*}
c-b&<-2\\
d-e&>1
\end{align*}
We have now:
\begin{align*}
g(-1) &= 1 + (c-b) - (d-e)\\
&<1-2-1\\
&<-2
\end{align*}
As $g(x)\to+\infty$ when $x\to\pm\infty$, the intermediate value theorem shows that $g(x)$ has at least one real root.

Alternate solution from Olinguito
Let
$$f(x)\ =\ ax^2+(c-b)x+(e-d)$$
and
$$g(x)\ =\ ax^4+bx^3+cx^2+dx+e\ =\ f(x^2)+b(x^3+x^2)+d(x+1).$$
If $f(r)=0$ then $r>1$ and
$$g\left(\sqrt r\right)\ =\ b(r\sqrt r+r)+d(\sqrt r+1)\ =\ (\sqrt r+1)(br+d)$$
and
$$g\left(-\sqrt r\right)\ =\ b(-r\sqrt r+r)+d(-\sqrt r+1)\ =\ (-\sqrt r+1)(br+d).$$
If $br+d=0$, then $g(x)=0$ has real roots $\pm\sqrt r$. Otherwise (if $br+d\ne0$) one of $g(\sqrt r),\,g(-\sqrt r)$ is positive and the other negative, since $\sqrt r+1>0$ and $-\sqrt r+1<0$; in this case $g(x)=0$ has a real root between $-\sqrt r$ and $\sqrt r$.

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