▶️ Ejercicio 271 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXIII

Ecuaciones incompletas de segundo grado

Ejercicio 271

Resolver las ecuaciones:

$\begin{matrix} 3 x^{2} & = 48 \\ x^{2} & = \frac{}{3} \\ x & = \sqrt{16} \\ x & = \pm 4 \end{matrix}$
$\begin{matrix} 5 x^{2} - 9 & = 46 \\ 5 x^{2} & = 46 + 9 \\ 5 x^{2} & = 55 \\ x^{2} & = \frac{}{5} \\ x & = \pm \sqrt{11} \end{matrix}$
$\begin{matrix} 7 x^{2} + 14 & = 0 \\ 7 (x^{2} + 2) & = 0 \\ x^{2} + 2 & = 0 \\ x^{2} & = - 2 \\ x & = \sqrt{- 2} \\ x & = \pm \sqrt{2} i \end{matrix}$
$\begin{matrix} 9 x^{2} - a^{2} & = 0 \\ (3 x + a) (3 x - a) & = 0 \\ 3 x + a & = 0 \\ 3 x & = - a \\ x_{1} & = - \frac{a}{3} \\ 3 x - a & = 0 \\ 3 x & = a \\ x_{2} & = \frac{a}{3} \end{matrix}$
$\begin{matrix} (x + 5) (x - 5) & = - 7 \\ x^{2} - 25 & = - 7 \\ x^{2} & = 25 - 7 \\ x^{2} & = 18 \\ x & = \sqrt{18} \\ x & = \sqrt{3^{2} .2} \\ x & = \pm 3 \sqrt{2} \end{matrix}$
$\begin{matrix} (2 x - 3) (2 x + 3) - 135 & = 0 \\ 4 x^{2} - 9 - 135 & = 0 \\ 4 x^{2} & = 144 \\ x^{2} & = \frac{}{4} \\ x & = \sqrt{36} \\ x & = \pm 6 \end{matrix}$
$\begin{matrix} 3 (x + 2) (x - 2) & = {(x - 4)}^{2} + 8 x \\ 3 (x^{2} - 4) & = x^{2} - 8 x + 16 + 8 x \\ 3 x^{2} - 12 & = x^{2} + 16 \\ 3 x^{2} - x^{2} & = 16 + 12 \\ 2 x^{2} & = 28 \\ x^{2} & = \frac{}{2} \\ x & = \pm \sqrt{14} \end{matrix}$
$\begin{matrix} (x + \frac{1}{3}) (x - \frac{1}{3}) & = \frac{1}{3} \\ x^{2} - \frac{1}{9} & = \frac{1}{3} \\ x^{2} & = \frac{1}{3} + \frac{1}{9} \\ x^{2} & = \frac{3 + 1}{9} \\ x^{2} & = \frac{4}{9} \\ x & = \sqrt{\frac{4}{9}} \\ x & = \pm \frac{2}{3} \end{matrix}$
$\begin{matrix} (2 x - 1) (x + 2) - (x + 4) (x - 1) + 5 & = 0 \\ 2 x^{2} + 4 x - x - 2 - (x^{2} + 3 x - 4) + 5 & = 0 \\ 2 x^{2} + 3 x - 2 - x^{2} - 3 x + 4 + 5 & = 0 \\ x^{2} + 7 & = 0 \\ x^{2} & = - 7 \\ x & = \sqrt{- 7} \\ x & = \pm \sqrt{7} i \end{matrix}$
$\begin{matrix} \frac{5}{2 x^{2}} - \frac{1}{6 x^{2}} & = \frac{7}{12} \\ \frac{15 - 1}{6 x^{2}} & = \frac{7}{} \\ 2 & = 7 x^{2} \\ x & = \sqrt{4} \\ x & = \pm 2 \end{matrix}$
$\begin{matrix} \frac{2 x - 3}{x - 3} & = \frac{x - 2}{x - 1} \\ (2 x - 3) (x - 1) & = (x - 2) (x - 3) \\ 2 x^{2} - 2 x - 3 x + 3 & = x^{2} - 5 x + 6 \\ 2 x^{2} - x^{2} & = 6 - 3 \\ x^{2} & = 3 \\ x & = \pm \sqrt{3} \end{matrix}$
$\begin{matrix} \frac{x^{2} - 5}{3} + \frac{4 x^{2} - 1}{5} - \frac{14 x^{2} - 1}{15} & = 0 \\ M u l t i p l i c a n d o l a e c u a c i ó n p o r 15 \\ 5 (x^{2} - 5) + 3 (4 x^{2} - 1) - 14 x^{2} + 1 & = 0 \\ 5 x^{2} - 25 + 12 x^{2} - 3 - 14 x^{2} + 1 & = 0 \\ 3 x^{2} - 27 & = 0 \\ 3 (x^{2} - 9) & = 0 \\ x^{2} - 9 & = 0 \\ x^{2} & = 9 \\ x & = \sqrt{9} \\ x & = \pm 3 \end{matrix}$
$\begin{matrix} 2 x - 3 - \frac{x^{2} + 1}{x - 2} & = - 7 \\ \frac{(2 x - 3) (x - 2) - x^{2} - 1}{x - 2} & = - 7 \\ 2 x^{2} - 4 x - 3 x + 6 - x^{2} - 1 & = - 7 (x - 2) \\ x^{2} - 7 x + 5 & = - 7 x + 14 \\ x^{2} & = - 5 + 14 \\ x^{2} & = 9 \\ x & = \sqrt{9} \\ x & = \pm 3 \end{matrix}$
$\begin{matrix} 3 - \frac{3}{4 x^{2} - 1} & = 2 \\ - \frac{3}{4 x^{2} - 1} & = 2 - 3 \\ \frac{3}{4 x^{2} - 1} & = 1 \\ 3 & = 4 x^{2} - 1 \\ 0 & = 4 x^{2} - 1 - 3 \\ 4 x^{2} - 4 & = 0 \\ 4 (x^{2} - 1) & = 0 \\ x^{2} - 1 & = 0 \\ (x - 1) (x + 1) & = 0 \\ x - 1 & = 0 \\ x_{1} & = 1 \\ x + 1 & = 0 \\ x_{2} & = - 1 \end{matrix}$

Categories: Capítulo XXXIII