Question

Aflati numarul numerelor irationale a pentru care numerele p=3a^2 -1 / a+1 si q = 2a^2+1 / 2a+3 sunt rationale

Answer 1

$Avem:~3a^2-1=ap+p \Rightarrow 6a^2-2=2ap+2p~~~~~(1) ~;\\ \\ 2a^2+1=2aq +3q \Rightarrow 6a^2+3=6aq+9q~~~~~(2).\\ \\ Scazand~relatia~(1)~din~relatia~(2),~obtinem: \\ \\ 5=6aq+9q-2ap-2p \Leftrightarrow 5-9q-2p=a(6q-2p) \Rightarrow a(6q-2p) \in Q. \\ \\ Dar~a \notin Q \Rightarrow 6q-2p=0 \ \textless \ =\ \textgreater \ 3q=p. \\ \\ Deci~ \frac{6a^2+3}{2a+3}= \frac{3a^2-1}{a+1} \Leftrightarrow 6a^3+6a^2+3a+3=6a^3+9a^2-2a-3 \Leftrightarrow \\ \\ \Leftrightarrow3a^2-5a-6=0$

$\Delta=97 \Rightarrow \boxed{a_1= \frac{5+ \sqrt{97}}{6}}~si~\boxed{a_2= \frac{5-\sqrt{97}}{6}}~. \\ \\ Ok...stiu~ca~rezolvarea~ecuatiei~de~gredul~2~este~materie~de~ \\ \\ semestrul~al~doilea,~asa~ca~iti~voi~prezenta~rezolvarea ~detaliata~a\\ \\ ecuatiei~ \boxed{3a^2-5a-6=0}~(*).\\ \\ (*) \Leftrightarrow (\sqrt{3}a)^2-2 \cdot(\sqrt{3}a) \cdot \frac{5}{2\sqrt{3}}+ \big(\frac{5}{2\sqrt{3}}\big)^2 -\big( \frac{5}{2\sqrt{3}} \big)^2-6=0 \Leftrightarrow$

$\Leftrightarrow~\big( \sqrt{3}a- \frac{5}{2 \sqrt{3}} \big)^2= \big(\frac{5}{2 \sqrt{3}} \big) ^2+6 \Rightarrow calcule....$