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Thechi1705
a fost răspuns

Sa se determine numarul real x astfel incat numerele reale x-2, [tex]\sqrt{5x+1}[/tex] , x+4 sa fie in progresie geometrica

Răspuns :

Marinalemandroi

[tex](\sqrt{5x+1})^2 =(x-2)(x+4)\\\\5x+1=x^2+2x-8\\\\x^2-3x-9=0\\\\\Delta=9+36=45\\\\x_1=\frac{3+3\sqrt{5} }{2} \\\\x_2=\frac{3-3\sqrt{5} }{2} \\\\Conditia\ de \ existenta\\5x+1\geq 0\\\\x\geq -\frac{1}{5}[/tex]

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