Question

3. Resolva, no conjunto R, as seguintes equações: a) x^{2}-2x=2x-4 b) x^{2}-2x=x+4 c) 6x^{2}+3x=1+2x d) 9x^{2}+3x+1=4x^{2}

Answer 1

Resposta:

3)

a) $x^{2}-2x=2x-4\\x^2-2x-2x+4=0\\x^2-4x+4=0\\\\\Delta=(-4)^2-4.1.4\\\Delta=16-16\\\Delta=0\\\\x=\frac{-(-4)\pm\sqrt{0}}{2}\\\\x'\hspace{2}=\hspace{2}x''\hspace{2}\\\\x=\frac{4}{2}\therefore\hspace{2}=2$          

 

b) $x^{2}-2x=x+4\\x^2-2x-x-4=0\\x^2-3x-4=0\\\\\Delta=(-3)^2-4.1.(-4)\\\Delta=9+16\therefore\hspace{2}\Delta=25\\\\x=\frac{-(-3)\pm\sqrt{25}}{2}\\x=\frac{3\pm5}{2}\\\\x'=\frac{3+5}{2}=\frac{8}{2}=4\hspace{50}x''=\frac{3-5}{2}=\frac{-2}{2}=-1$

c) $6x^{2}+3x=1+2x\\6x^2+x-1=0\\\\\Delta=1^2-4.6.(-1)\\\Delta=1+24\hspace{2}\therefore\hspace{2}\Delta=25\\\\x=\frac{-1\pm\sqrt{25}}{12}\\x=\frac{-1\pm5}{12}\\\\x'=\frac{-1+5}{12}=\frac{4}{12}=\frac{1}{3}\hspace{25}x''=\frac{-1-5}{12}=\frac{-6}{12}=-\frac{1}{2}$

d) $9x^{2}+3x+1=4x^{2}\\5x^2+3x+1=0\\\\\Delta=3^2-4(5).1\\\Delta=9-20\\\Delta=-11\\\\S\not\in\mathbb{R}$