Question

URGENTT !! Clasa aV-a!! Coroana si 13 puncte.
a)X supra 3 din X+1 este 4;
b)X+1 supra X+2 din 2*X+4 este 12

Answer 1

$\displaystyle a) \frac{x}{3} \cdot x+1=4 \Rightarrow \frac{x(x+1)}{3} =4 \Rightarrow \frac{x^2+x}{3} =4 \Rightarrow x^2+x=12 \Leftrightarrow \\ \\ \Leftrightarrow x^2+x-12=0 \\ \\ a=1,b=1,c=-12 \\ \\ D=b^2-4ac=1^2-4 \cdot 1 \cdot (-12)=1+48=49\ \textgreater \ 0 \\ \\ x_1= \frac{-1+ \sqrt{49} }{2 \cdot 1} = \frac{-1+7}{2} = \frac{6}{2} =3 \\ \\ x_2= \frac{-1- \sqrt{49} }{2 \cdot 1} = \frac{-1-7}{2 } = \frac{-8}{2} =-4$

$\displaystyle b) \frac{x+1}{x+2} \cdot 2x+4=12 \Rightarrow \frac{(x+1)(2x+4)}{x+2} =12 \Leftrightarrow \\ \\ \Leftrightarrow \frac{2x^2+4x+2x+4}{x+2} =12 \Rightarrow \frac{2x^2+6x+4}{x+2} =12 \Leftrightarrow \\ \\ \Leftrightarrow 2x^2+6x+4=12(x+2) \Rightarrow 2x^2+6x+4=12x+24 \Leftrightarrow \\ \\ \Leftrightarrow2x^2+6x-12x=24-4 \Rightarrow 2x^2-6x=20 \Leftrightarrow \\ \\ \Leftrightarrow 2x^2-6x-20=0 \\ \\ a=2,b=-6,c=-20$

$\displaystyle D=b^2-4ac=6^2-4 \cdot 2 \cdot (-20)=36+160=196\ \textgreater \ 0 \\ \\ x_1= \frac{6+ \sqrt{196} }{2 \cdot 2} = \frac{6+14}{4} = \frac{20}{4} =5 \\ \\ x_2= \frac{6- \sqrt{196} }{2 \cdot 2} = \frac{6-14}{4} = \frac{-8}{4} =-2$