Îmi LIPSEȘTI . Mi-e DOR de TINE // Problem C:34

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"C:34.  Consider polynomials  

$P(x)=2x^2+ax+b\; and\;Q(x)=x^2+cx+d\;;\;a,\;b,\;c,\;d\in\mathbb{R}$ 

 such that  $a^2+c^2+4d\leqslant 2ac+4b$. Show that if  $P(x)$  has real roots,

             then  $Q(x)$  also has real roots. Is the converse true?

{Author : } Costache CIOTLAȘ, teacher, Comănești, Bacău"

ANSWER CiP

The converse statement is NOT true : $a=3,\;b=3,\;c=3,\;d=2$ 

 satisfies the condition  $a^2+c^2+4d=26<30=2ac+4b$ , $Q(x)=x^2+3x+2$  has the roots  $-1\; and\;-2$ ,

 but  $P(x)=2x^2+3x+3$  has no real roots, as its discriminant is $-15<0.$

                                       Solution CiP

               If  $P(x)$  has real roots, then its discriminant is

$\Delta_P=a^2-4\cdot 2\cdot b=a^2-8b\geqslant 0 \tag{1}$

Polynomial  $Q(x)$  has discriminant  $\Delta_Q=c^2-4d$.

But  $-4d\geqslant a^2+c^2-2ac-4b\;\Rightarrow \Delta_Q\geqslant c^2+a^2+c^2-2ac-4b=$

$=a^2-2ac+2c^2-4b=\left (\frac{a^2}{2}-2ac+2c^2 \right )+\left ( \frac{a^2}{2}-4b\right )=$

$=\frac{(a-2c)^2}{2}+\frac{a^2-8}{2}=\frac{(a-2c)^2}{2}+\frac{\Delta_P}{2}\underset{(1)}{\geqslant} 0$

so  $Q(x)$  has real roots too.

$\blacksquare$