Question

12 și 13 vă rooog!!!

Answer 1

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

$\displaystyle b).(x-3)^2+2(x+1)=7 \\ \\ x^2-2 \cdot x \cdot 3+3^2+2x+2=7 \\ \\ x^2-6x+9+2x+2=7 \\ \\ x^2-6x+2x+9+2-7=0 \\ \\ x^2-4x+4=0 \\ \\ a=1,~ b=-4,~ c=4 \\ \\ \Delta=(-4)^2-4 \cdot 1 \cdot 4=16-16=0 \\ \\ \Delta=0 \Rightarrow x_1=x_2=- \frac{b}{2a} \\ \\ x_1=x_2= \frac{4}{2} =2$

$\displaystyle c). \frac{x^2+5}{2} = \frac{2x+13}{5} \\ \\ 5(x^2+5)=(2x+13) \cdot 2 \\ \\ 5x^2+25=4x+26 \\ \\ 5x^2-4x+25-26=0 \\ \\ 5x^2-4x-1=0 \\ \\ a=5,~b=-4,~c=-1 \\ \\ \Delta=(-4)^2-4 \cdot 5 \cdot (-1)=16+20=36\ \textgreater \ 0 \\ \\ x_1= \frac{4-6}{10} =- \frac{2}{10} =- \frac{1}{5} \\ \\ x_2= \frac{4+6}{10} = \frac{10}{10} =1$

$\displaystyle d). \frac{x+2}{x+8} = \frac{2-x}{x-8} \\ \\ (x+2)(x-8)=(2-x)(x+8) \\ \\ x^2-8x+2x-16=2x+16-x^2-8x \\ \\ x^2-8x+2x-16-2x-16+x^2+8x=0 \\ \\ 2x^2-32=0 \Rightarrow 2x^2=32 \Rightarrow x^2= \frac{32}{2} \Rightarrow x^2=16 \Rightarrow x=4 \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-4$

$\displaystyle 13a). |x|= \frac{1}{3} \Rightarrow x= \frac{1}{3} \\ \\ ~~~~~~~~~~~~~~~~~~~~~~x=- \frac{1}{3}$

$\displaystyle b).2|x|= \frac{5}{6} \Rightarrow |x|= \frac{5}{12} \Rightarrow x= \frac{5}{12} \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=- \frac{5}{12}$

$\displaystyle c).|2x-1|= \frac{1}{4} \\ \\ 2x-1= \frac{1}{4} \Rightarrow 8x-4=1 \Rightarrow 8x=1+4 \Rightarrow 8x=5 \Rightarrow x= \frac{5}{8} \\ \\ 2x-1=- \frac{1}{4} \Rightarrow 8x-4=-1 \Rightarrow 8x=-1+4 \Rightarrow 8x=3 \Rightarrow x= \frac{3}{8}$