Question

$6x {}^{2} + \frac{5}{2}x + \frac{1}{4} = 0$
Por factorización

Answer 1

$\displaystyle 6x^2+\frac{5}{2}x+\frac{1}{4}=0 \\[2pt] \text{Lo recomendable es trabajar con enteros , para ellos se multiplica}\\ \text{por 4 ambos lados de la igualdad.} \\[2pt] 4\left( 6x^2+\frac{5}{2}x+\frac{1}{4} \right)=(4)(0) \\[6pt] 24x^2+10x+1=0 \\ 6x \hspace*{1.75cm}1 \\ 4x \hspace*{1.75cm}1 \\[6pt] (6x+1)(4x+1)=0 \\[2pt] \Rightarrow \ 6x+1=0 \ \vee \ 4x+1=0 \\[2pt] \Rightarrow \ \ x=-\frac{1}{6} \hspace*{0.3cm}\ \vee \ \ x=-\frac{1}{4}$

$\text{Las soluciones son : } \\[4pt] \displaystyle x=\left \{ -\frac{1}{4} \ , \ -\frac{1}{6} \right \}$

Answer 2

▪Enunciado
$\boxed{6 {x}^{2} + \frac{5}{2} x + \frac{1}{4} = 0}$

▪Procedimiento

$6 {x}^{2} + \frac{5}{2} x + \frac{1}{4} = 0 \: \: \: \: \: (8) \\ \\ (8)6{x}^{2} + ( 8) \frac{5}{2} x + (8) \frac{1}{4} = (8)0 \\ \\ 48 {x}^{2} + 20x + 2 = 0 \\ \\ (6x + 1)(8x + 2)$

° Dos posibles soluciones:
$x _{1} \begin{cases}6x + 1 = 0 \\ 6x = - 1 \\ \boxed{ \boxed{x = - \frac{1}{6} }} \end{cases} \\ \\ x _{2} \begin{cases} 8x + 2 = 0 \\ 8x = - 2 \\ x = - \frac{2}{8} \\ \boxed{ \boxed{ x = - \frac{1}{4} }} \end{cases}$