Question

va rogg exercitiile 4,5 si 6
le vreau pe toate trei revolvate si dau cel mai bun raspuns

Answer 1

4.

[tex]a.\\\\ x^2-2x-1=0\\ \Delta=b^2-4*a*c\\ \Delta=(-2)^2-4*1*(-1)\\ \Delta=4+4\\ \Delta=8\\\\ \bold{ \sqrt{\Delta} = \sqrt{8}=2 \sqrt{2}}\\ \\ x_1= \frac{-b+\sqrt{\Delta} }{2*a} = \frac{-(-2)+2 \sqrt{2} }{2} = \frac{2+2 \sqrt{2} }{2}= \frac{2*(1+ \sqrt{2} }{2}\bold{=1+ \sqrt{2} }\\\\ x_2= \frac{-b-\sqrt{\Delta} }{2*a} = \frac{-(-2)-2 \sqrt{2} }{2} = \frac{2-2 \sqrt{2} }{2}= \frac{2*(1-\sqrt{2} }{2}\bold{=1- \sqrt{2} }\\\\\\ b~,~c~si~d~sunt~foarte asemanatoare~ cu~ punctul ~a. \uparrow~ \uparrow~ \uparrow\\\\ [/tex]

[tex]e.\\x^2-4=0\\ (x-2)(x+2)=0~\Longrightarrow~ \left {{(x-2)=0~\Longrightarrow~\bold{x=2}} \atop {(x+2)=0 ~\Longrightarrow~\bold{x=-2}} \right. [/tex]

[tex]f. \\ 2x^2+3x=0~\Longrightarrow~x*(2x+3)=0~\Longrightarrow~ \left {{x=0} \atop {2x+3=0 \Longrightarrow2x=-3 \Longrightarrow~x= -\frac{3}{2} } \right. [/tex]


[tex]5.\\ a.\\\\x^2-1=x-3\\ x^2-x-1+3=0\\ x^2-x+2=0\\ \Delta=b^2-4*a*c=(-1)^2-4*1*2=1-8=-7\\\Longrightarrow~Nu ~esista~solutii!!!\\\\ b.\\ 2(x+1)-x(x+1)=0\\ 2x+2-x^2-1=0\\ -x^2+2x+1=0\\ \Delta=b^2-4*a*c=2^2-4*(-1)*1=4+4=8\\ \sqrt{\Delta}=\sqrt{8}= 2\sqrt{2} \\\\ x_1= \frac{-b+\sqrt{\Delta}}{2*a}= \frac{-2+2 \sqrt{2}}{-2}= \frac{-2*(1- \sqrt{2}}{-2}=\bold{1- \sqrt{2}} \\\\ x_2= \frac{-b-\sqrt{\Delta}}{2*a}= \frac{-2-2 \sqrt{2}}{-2}= \frac{-2*(1+ \sqrt{2}}{-2}=\bold{1+ \sqrt{2}} \\\\ [/tex]


[tex]c.\\\\ x^2-6x+9=4\\ x^2-6x+9-4=0\\ x^2-6x+5=0\\ \Delta=b^2-4*a*c=36-4*1*5=36-20=16\\ \sqrt{\Delta} = \sqrt{16}=4\\\\ x_1= \frac{-b+\sqrt{\Delta}}{2*a}= \frac{-(-6)+4}{2} = \frac{6+4}{2}= \frac{10}{2} \bold{=5}\\\\ x_2= \frac{-b-\sqrt{\Delta}}{2*a}= \frac{-(-6)-4}{2} = \frac{6-4}{2}= \frac{2}{2} \bold{=1}\\\\\\\\ [/tex]

[tex]d. \\ \\\frac{x^2-1}{2} = \frac{x+1}{3}\\ 2*(x+1)=3*(x^2-1)\\ 2x+1=2x^2-3\\ -2x^2+2x+1+3=0\\ -2x^2+2x+4=0\\ \Delta=b^2-4*a*c=4+32=36\\ \sqrt{\Delta}= \sqrt{36}=6 \\\\ x_1= \frac{-b+\sqrt{\Delta}}{2*a}= \frac{-2+6}{2*(-2)} = \frac{4}{-4} =\bold{-1}\\ x_2=\frac{-b-\sqrt{\Delta}}{2*a}= \frac{-2-6}{2*(-2)} = \frac{-8}{-4} =\bold{2}\\\\ e.\\ 1-(x-2)^2=0\\ 1-x^2+2x-4=0\\ -x^2+2x-3=0\\ \Delta=b^2-4*a*c=4-12=-8\\ =\ \textgreater \ Ecuatia~nu~are~solutie~deoarece~\bold{\Delta\ \textless \ 0~}\\[/tex]