Proszę o rozwiązanie analizy metodą starożytnych [latex]3- sqrt{x-1}=sqrt{3x-2}[/latex]

[latex] 3- sqrt{x-1}=sqrt{3x-2} /()^2[/latex] [latex] 9- 6sqrt{x-1}+x-1=3x-2[/latex] [latex] - 6sqrt{x-1}=3x-2-9-x+1[/latex] [latex] - 6sqrt{x-1}=2x-10 /()^2[/latex] [latex]36(x-1)=4x^2-40x+100[/latex] [latex]36x-36=4x^2-40x+100[/latex] [latex]4x^2-40x+100-36x+36=0[/latex] [latex]4x^2-76x+136=0 /:4[/latex] [latex]x^2-19x+34=0[/latex] [latex]Delta=(-19)^2-4 cdot 1 cdot 34=361-136=225[/latex] [latex] sqrt{Delta}= sqrt{225} =15[/latex] [latex]x_1= frac{19-15}{2} = frac{4}{2}=2 [/latex] [latex]x_2= frac{19+15}{2} = frac{34}{2}=17[/latex] --------------------- [latex]x=2[/latex] [latex]L=3- sqrt{x-1}=3- sqrt{2-1}=3- sqrt{1}=3-1=2[/latex] [latex]P=sqrt{3x-2}=sqrt{3 cdot 2-2}=sqrt{6-2}= sqrt{4} =2[/latex] [latex]L=P[/latex] --------------------- [latex]x=17[/latex] [latex]L=3- sqrt{x-1}=3- sqrt{17-1}=3- sqrt{16}=3-4=-1[/latex] [latex]P=sqrt{3x-2}=sqrt{3 cdot 17-2}=sqrt{51-2}= sqrt{49} =7[/latex] [latex]L eq P[/latex] --------------------- Rozwiązaniem równania jest x=2