Question

SA se afle valorile parametrului real a pentru care o solutie a ecuatiei 4xpatrat-15x+4acub=0 este patratul celeilalte solutii.

Answer 1

[tex]4x^2-15x+4a^3=0 |:4\\ x^2-\dfrac{15}{4}\cdot x+a^3=0\\ \text{Mai intai punem conditia ca } \Delta \text{ sa fie mai mare ca 0:}\\ \Delta=\dfrac{225}{16}-4a^3 \ \textgreater \ 0\\ \\ ~~~~~~~\dfrac{225}{16} \ \textgreater \ 4a^3\\ \\ ~~~~~~~a^3 \ \textless \ \dfrac{225}{64}\\ ~~~~~~~a\ \textless \ \dfrac{\sqrt[3]{225}}{4}\\ [/tex]
[tex]\text{Fie}\ x_1 \text{ si}\ x_2\ \text{cele doua solutii ale ecuatiei.Folosind relatiile lui Viete} \\ \text{vom avea:}\\ P=x_1\cdot x_2= a^3\\ S=x_1+x_2=\dfrac{15}{4}\\ \text{Din relatia din enunt avem ca :} x_2=x_1^2\\ \text{Am obtinut un sistem de 3 relatii cu 3 necunoscute,deci este posibil}\\ \text{de rezolvat:}\\ x_1+x_1^2=\dfrac{15}{4}\\ x_1^2+x_1-\dfrac{15}{4}=0\\ \Delta=1+15=16\Rightarrow \sqrt{\Delta}=4\\ x_1=\dfrac{-1+4}{2} =\dfrac{3}{2}\\ \bold{sau}\ x_1=\dfrac{-1-4}{2}=\dfrac{-5}{2}\\ [/tex]
[tex]\text{Pentru}\ x_1=\dfrac{3}{2} : \\ x_1 ^ 3=a^3\Rightarrow a=\dfrac{27}{8} \ \textgreater \ \dfrac{\sqrt[3]{225}}{4}\\ \text{ Pentru}\ x_1=-\dfrac{5}{2}:\\ x_1^3=a^3\Rightarrow a=\dfrac{-125}{8} \ \textless \ \dfrac{\sqrt[3]{225}}{4}\\ \text{Singura solutie valabila este: } \boxed{a=-\dfrac{125}{8}}[/tex]