Question

resolva as equações utilizando o método de completar quadrados: a) 4x^2+12x+5=0; b) x^2+10x/3+1=0

Answer 1

a) $4x^{2}+12x+5=0$

$4x^{2}+12x=-5\\ \\ 4x^{2}+12x+9=-5+9\\ \\ (2x+3)^{2}=4\\ \\ 2x+3=\pm \sqrt{4}\\ \\ 2x+3=\pm 2\\ \\ 2x=\pm 2-3\\ \\ x=\dfrac{\pm 2-3}{2}\\ \\ \\ \begin{array}{rcl} x=\dfrac{2-3}{2}&\text{ ou }&x=\dfrac{-2-3}{2}\\ \\ x=-\dfrac{1}{2}&\text{ ou }&x=-\dfrac{5}{2} \end{array}$


b) $x^{2}+\dfrac{10x}{3}+1=0$

$x^{2}+\dfrac{10x}{3}=-1\\ \\ \\ x^{2}+\dfrac{10x}{3}+\dfrac{25}{9}=-1+\dfrac{25}{9}\\ \\ \\ \left(x+\dfrac{5}{3} \right )^{2}=\dfrac{-9+25}{9}\\ \\ \\ \left(x+\dfrac{5}{3} \right )^{2}=\dfrac{16}{9}\\ \\ \\ x+\dfrac{5}{3}=\pm \sqrt{\dfrac{16}{9}}\\ \\ \\ x+\dfrac{5}{3}=\pm \dfrac{4}{3}\\ \\ \\ x=\pm \dfrac{4}{3}-\dfrac{5}{3}\\ \\ \\ \begin{array}{rcl} x=\dfrac{4}{3}-\dfrac{5}{3}&\text{ ou }&x=-\dfrac{4}{3}-\dfrac{5}{3}\\ \\ \\ x=-\dfrac{1}{3}&\text{ ou }&x=-\dfrac{9}{3}\\ \\ \\ x=-\dfrac{1}{3}&\text{ ou }&x=-3 \end{array}$