Question

Quais as raízes reais da equação (x + 5)² - (2x+1)² = 2 . Demonstre o calculo.

Answer 1

$(x^2+10x+25)-(4x^2+4x+1)=2\\ \\ -3x^2+6x+24=2\\\\ -3x^2+6x+22=0\\\\ 3x^2-6x-22=0\\\\ x_{1,2}=\dfrac{6\pm\sqrt{36-4(3)(-22)}}{6}\\ \\ x_{1,2}=\dfrac{6\pm\sqrt{300}}{6}$

                          $\boxed{x_{1,2}=\dfrac{3\pm5\sqrt{3}}{3}}$

Answer 2

$(x + 5)^2 - (2x+1)^2= 2\to \\\\ ( x^{2} +10x+25)-(4 x^{2} +4x+1)=2\to \\\\ x^{2} +10x+25-4 x^{2} -4x-1-2=0\to \\\\ -3 x^{2} +6x+22=0\\\\ a=-3;\ b=6;\ c=22\\\\\\ \Delta=b^2-4ac\to \Delta=6^2-4.(-3).22\to \Delta=36+264\to \boxed{\Delta=300}\\\\ x' \neq x''$



$x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \to ~~ x= \dfrac{-6\pm \sqrt{3'00} }{2.(-3)} \to ~~ x= \dfrac{-6\pm 10\sqrt{3} }{-6} \to \\\\\\ x'= \dfrac{-6+ 10\sqrt{3} }{-6} \to ~~x'= \dfrac{-3+ 5\sqrt{3} }{-3} \to ~~ \boxed{x'= \dfrac{3-5\sqrt{3} }{3} }\\\\\\ x'= \dfrac{-6-10\sqrt{3} }{-6} \to ~~x'= \dfrac{-3- 5\sqrt{3} }{-3} \to ~~ \boxed{x'= \dfrac{3+5\sqrt{3} }{3} }\\\\\\\\\\ \boxed{\boxed{S=\{ \dfrac{3-5\sqrt{3} }{3} ~~;~~ \dfrac{3+5\sqrt{3} }{3} \}}}$