Question 1121268
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You asked: Solve by completing the perfect square trinomial


We must have a language problem here because your quadratic is not a perfect square trinomial.  I'm almost certain that you wanted to ask "Solve 2x^2 - 5x - 12 = 0 by completing the square"


<i>Aquí debemos tener un problema de lenguaje porque su cuadrática no es un trinomio cuadrado perfecto. Estoy casi seguro de que querías preguntar "Resuelve 2x ^ 2 - 5x - 12 = 0 completando el cuadrado"</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 5x\ -\ 12\ =\ 0]


Add the additive inverse of the constant term to both sides and then divide both sides by the lead coefficient:


<i>Agregue el aditivo inverso del término constante a ambos lados y luego divida ambos lados por el coeficiente principal:</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{5}{2}x\ =\ 6]


Divide the first-degree coefficient by 2, square the result, and add that result to both sides of the equation.


<i>Divida el coeficiente de primer grado entre 2, cuadre el resultado y agregue ese resultado a ambos lados de la ecuación.</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{5}{2}x\ +\ \frac{25}{16}\ =\ 6\ +\ \frac{25}{16}]


The left-hand side is now a perfect square.  Factor it and simplify the right-hand side:


<i> El lado izquierdo ahora es un cuadrado perfecto. Factorícelo y simplifique el lado derecho:</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(x\ -\ \frac{5}{4}\)^2\ =\ \frac{121}{16}]


Take the square root remembering to include both positive and negative roots


<i>Tome la raíz cuadrada recordando incluir raíces positivas y negativas</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{5}{4}\ =\ \pm\sqrt{\frac{121}{16}}\ =\ \pm\frac{11}{4}]


Then


<i>Entonces</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{11}{4}\ +\ \frac{5}{4}\ =\ 4]


or


<i>o</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{11}{4}\ +\ \frac{5}{4}\ =\ \frac{3}{2}]



								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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