Question

Sa se detrmine valorile reale ale numarului x , stiind ca numerele 5 - x , x + 7 si 3x + 11 sunt termeni consecutivi ai unei progresii geometrice .

Answer 1

$x+7= \sqrt{(5-x)(3x+11)} \\ x^{2} +49+14x=15x+55-3 x^{2} -11x \\ 4x^{2} +10x-6=0 \\ 2 x^{2} +5x-3=0 \\ delta=49 \\ x1=1/2 \\ x2=-3 \\ dar x apartine[-11/3.5] deci sunt solutii$

Answer 2

$\displaystyle 5-x,~x+7,~3x+11 \\ \\ (x+7)^2=(5-x)(3x+11) \\ \\ x^2+2 \cdot x \cdot 7+7^2=15x+55-3x^2-11x \\ \\ x^2+14x+49=15x+55-3x^2-11x \\ \\ x^2+3x^2+14x-15x+11x=55-49 \\ \\ 4x^2+10x=6 \\ \\ 4x^2+10x-6=0|:2 \\ \\ 2x^2+5x-3=0 \\ \\ a=2,~b=5,~c=-3 \\ \\ \Delta=b^2-4ac=5^2-4 \cdot 2 \cdot (-3)=25+24=49\ \textgreater \ 0 \\ \\ x_1= \frac{-5+ \sqrt{49} }{2 \cdot 2} = \frac{-5+7}{4} = \frac{2}{4} = \frac{1}{2} \\ \\ x_2= \frac{-5- \sqrt{49} }{2 \cdot 2} = \frac{-5-7}{4} = \frac{-12}{4} =-3$