Let $P$ be the polynomial defined by $P (Z) = Z^{4}-aZ^{2} +1$
Calculate $P\left(\sqrt{\dfrac{a}{2}}\right)$ and deduce the number of real roots of $P$ as the case $a = 2$, $a >2$ or $a <2$.
My attempt:
- First Method
$P\left(\sqrt{\dfrac{a}{2}}\right)=\dfrac{a^{2}}{4}-a\dfrac{|a|}{2}+1$
- Case $a=2$
$\Longrightarrow P\left(\sqrt{\dfrac{a}{2}}\right)=0 $ then the polynomial $P$ has solution $\sqrt{\dfrac{a}{2}}=1$
as $1$ is root of $P$ then by Euclidean division the polynomial has the following factors: $$P(Z)=(Z-1).(Z^{3}+2^{2}-2+1)$$
note that $(Z^{3}+2^{2}-2+1)$ has trivial roots $1$ and $-1$ then by Euclidean division
\begin{align} P(Z)&=(Z-1)^{2}.(Z^{2}+2Z+1)\\ &=(Z-1)^{2}.(Z+1)^{2} \end{align}
Then $P$ has multiple roots $-1$ and $1$ .
- Case $a>2$
\begin{align} P\left(\sqrt{\dfrac{a}{2}}\right)&=\dfrac{a^{2}}{4}-a\dfrac{|a|}{2}+1\\ &=1-\dfrac{a^{2}}{4} \end{align}
let $Y=Z^{2}$ then $P(Z)=Q(Y)=Y^{2}-aY+1$
the discriminant of $Q$ is $\Delta=a^{2}-4$ as $a>2 \Longrightarrow \Delta >0$ then $Q$ has two real roots $Y=\left( \dfrac{a \pm \sqrt{a^{2}-4}}{2}\right)$ thus $P$ has two real roots $Z^{2}=\left( \dfrac{a \pm \sqrt{a^{2}-4}}{2}\right) \Longrightarrow $
- Case $a<2$
let $Y=Z^{2}$ then $P(Z)=Q(Y)=Y^{2}-aY+1$ as $a<2$ then $\Delta<0$ has no real roots
- Second Method
Using Intermediate value theorem
- Case $a>2$
$P'(Z)=4Z^{3}-2.aZ$ then if $P'(Z)=0 \Longrightarrow Z=0 \text{ or } Z=\pm \sqrt{\dfrac{a}{2}}$
on $[0, \sqrt{\dfrac{a}{2}}]$
since $P$ is monotone and continuous on $[0, \sqrt{\dfrac{a}{2}}]$ then by Intermediate value theorem Then there exists $Z$ between $0$ and $\sqrt{\dfrac{a}{2}}$ such that $P(Z)=c$.
on $[\sqrt{\dfrac{a}{2}},+\infty]$
since $P$ is monotone and continuous on $[0, \sqrt{\dfrac{a}{2}}]$ then by Intermediate value theorem Then there exists $Z$ between $+\infty$ and $\sqrt{\dfrac{a}{2}}$ such that $P(Z)=c$.
Case $a<2$
there is solution just in case of $0<a<2$ and its the same as above
- My reasoning, is it correct ?
- Is there another simple ways to solve that question ?
- I add the new follwing method:
$P(Z)=Z^{4}-aZ^{2}+1=(Z^{2}-\dfrac{a}{2})^2-\dfrac{a^{2}-4}{4}$
immediately gives the results :
- for case a=2
$ P(Z)=(Z-1)^{2}(Z+1)^{2}$ then p has two multiple solution on $\mathbb{R}$
- For Case $|a|<2$
then $P(Z)$ is a sum of two squares, one of which is not zero so $P(z) > 0$ then P has non roots on $\mathbb{R}$
- For Case $|a|>2$
the canonical form of $P(z)$ is a difference of squares so we can factor as :
$$P(Z)=(z-\dfrac{a-\sqrt{a^{2}-4}}{2})(z-\dfrac{a+\sqrt{a^{2}-4}}{2})$$
then $P$ has two solution on $\mathbb{R}$
- For Case $a=-2$
we've $P(Z)=(Z^{2}+1)^{2}\geq 1$ then p has no solution on $\mathbb{R}$
Yes, this is essentially the way to go. You could have obtained the factorization in the case $a=2$ as follows :
Since $$ P(z) = z^4 - 2z^2 + 1 = (z^2)^2 - 2(z^2) + 1 = q(z^2) $$ where $q(z) = z^2 - 2z + 1 = (z-1)^2$ (if you don't see this, the rational root theorem tells you the only possible roots of $q$ are $\pm 1$, so you can check if they are and then use the Euclidean algorithm or write $z = z-1+1$, substitute and expand). Therefore $p(z) = (z^2-1)^2 = (z-1)^2 (z+1)^2$.
This is better in some sense because you don't have to factor a quartic by hand.
Hope that helps,