Question

lg(x-5)+lg(x+4)=lg2+lg5

Answer 1

DVA: X>5
(x-5)(x+4)=10
x^2-x-30=0
Δ=121
x=-5;x=6
asa cum -5 nu poate fi solutie,rezulta ca x=6 este raspunsul
S={6}

Answer 2

$\displaystyle \mathtt{lg(x-5)+lg(x+4)=lg2+lg5~~~~~~~~~~~~~~~~~~~~~~~~~~C.E. \left \{ {{x-5\ \textgreater \ 0} \atop {x+4\ \textgreater \ 0}} \right. }~\\ \\ \mathtt{lg((x-5)(x+4))=lg(2 \cdot 5)}\\ \\ \mathtt{lg((x-5)(x+4))=lg10}\\ \\ \mathtt{(x-5)(x+4)=10}\\ \\ \mathtt{x^2+4x-5x-20=10}\\ \\ \mathtt{x^2-x-20-10=0}\\ \\ \mathtt{x^2-x-30=0}\\ \\ \mathtt{a=1,~b=-1,~c=-30}\\ \\ \mathtt{\Delta=b^2-4ac=(-1)^2-4 \cdot 1 \cdot (-30)=1+120=121\ \textgreater \ 0}$
$\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-1)+ \sqrt{121} }{2 \cdot 1}= \frac{1+11}{2}= \frac{12}{2}=6 } \\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-1)- \sqrt{121} }{2 \cdot 1}= \frac{1-11}{2}= \frac{-10}{2}=-5}\\ \\ \mathtt{x_1=6 \Rightarrow \left \{ {{6-5\ \textgreater \ 0~A} \atop {6+4\ \textgreater \ 0~A}} \right.\Rightarrow x_1=6~este~solutie~a~ecuatiei. } \\ \\ \mathtt{x_2=-5 \Rightarrow -5-5\ \textgreater \ 0~F\Rightarrow x_2=-5~nu~este~solutie~a~ecuatiei.}\\ \\ \mathtt{S=\{6\}}$