Sr Examen

Otras calculadoras

|x+3|=a la ecuación

El profesor se sorprenderá mucho al ver tu solución correcta😉

v

Solución numérica:

Buscar la solución numérica en el intervalo [, ]

Solución

Ha introducido [src] 
|x + 3| = a
∣x+3∣=a\left|{x + 3}\right| = a
Simplificar
Solución detallada
Para cada expresión dentro del módulo en la ecuación
admitimos los casos cuando la expresión correspondiente es ">= 0" o "< 0",
resolvemos las ecuaciones obtenidas.

1.
x+3≥0x + 3 \geq 0
o
−3≤x∧x<∞-3 \leq x \wedge x < \infty
obtenemos la ecuación
−a+(x+3)=0- a + \left(x + 3\right) = 0
simplificamos, obtenemos
−a+x+3=0- a + x + 3 = 0
la resolución en este intervalo:
x1=a−3x_{1} = a - 3

2.
x+3<0x + 3 < 0
o
−∞<x∧x<−3-\infty < x \wedge x < -3
obtenemos la ecuación
−a+(−x−3)=0- a + \left(- x - 3\right) = 0
simplificamos, obtenemos
−a−x−3=0- a - x - 3 = 0
la resolución en este intervalo:
x2=−a−3x_{2} = - a - 3


Entonces la respuesta definitiva es:
x1=a−3x_{1} = a - 3
x2=−a−3x_{2} = - a - 3
Gráfica
Suma y producto de raíces [src] 
suma
    //-3 - a  for a > 0\     //-3 - a  for a > 0\       //-3 + a  for a >= 0\     //-3 + a  for a >= 0\
I*im|<                 | + re|<                 | + I*im|<                  | + re|<                  |
    \\ nan    otherwise/     \\ nan    otherwise/       \\ nan    otherwise /     \\ nan    otherwise /
(re⁡({−a−3for a>0NaNotherwise)+iim⁡({−a−3for a>0NaNotherwise))+(re⁡({a−3for a≥0NaNotherwise)+iim⁡({a−3for a≥0NaNotherwise))\left(\operatorname{re}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}\right) + \left(\operatorname{re}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}\right) 
=
    //-3 + a  for a >= 0\       //-3 - a  for a > 0\     //-3 + a  for a >= 0\     //-3 - a  for a > 0\
I*im|<                  | + I*im|<                 | + re|<                  | + re|<                 |
    \\ nan    otherwise /       \\ nan    otherwise/     \\ nan    otherwise /     \\ nan    otherwise/
re⁡({−a−3for a>0NaNotherwise)+re⁡({a−3for a≥0NaNotherwise)+iim⁡({−a−3for a>0NaNotherwise)+iim⁡({a−3for a≥0NaNotherwise)\operatorname{re}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + \operatorname{re}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
producto
/    //-3 - a  for a > 0\     //-3 - a  for a > 0\\ /    //-3 + a  for a >= 0\     //-3 + a  for a >= 0\\
|I*im|<                 | + re|<                 ||*|I*im|<                  | + re|<                  ||
\    \\ nan    otherwise/     \\ nan    otherwise// \    \\ nan    otherwise /     \\ nan    otherwise //
(re⁡({−a−3for a>0NaNotherwise)+iim⁡({−a−3for a>0NaNotherwise))(re⁡({a−3for a≥0NaNotherwise)+iim⁡({a−3for a≥0NaNotherwise))\left(\operatorname{re}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}\right) \left(\operatorname{re}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}\right) 
=
/-(-3 + I*im(a) + re(a))*(3 + I*im(a) + re(a))  for a > 0
<                                                        
\                     nan                       otherwise
{−(re⁡(a)+iim⁡(a)−3)(re⁡(a)+iim⁡(a)+3)for a>0NaNotherwise\begin{cases} - \left(\operatorname{re}{\left(a\right)} + i \operatorname{im}{\left(a\right)} - 3\right) \left(\operatorname{re}{\left(a\right)} + i \operatorname{im}{\left(a\right)} + 3\right) & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases} 
Piecewise((-(-3 + i*im(a) + re(a))*(3 + i*im(a) + re(a)), a > 0), (nan, True))
Respuesta rápida [src] 
         //-3 - a  for a > 0\     //-3 - a  for a > 0\
x1 = I*im|<                 | + re|<                 |
         \\ nan    otherwise/     \\ nan    otherwise/
x1=re⁡({−a−3for a>0NaNotherwise)+iim⁡({−a−3for a>0NaNotherwise)x_{1} = \operatorname{re}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - a - 3 & \text{for}\: a > 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
         //-3 + a  for a >= 0\     //-3 + a  for a >= 0\
x2 = I*im|<                  | + re|<                  |
         \\ nan    otherwise /     \\ nan    otherwise /
x2=re⁡({a−3for a≥0NaNotherwise)+iim⁡({a−3for a≥0NaNotherwise)x_{2} = \operatorname{re}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} a - 3 & \text{for}\: a \geq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
Eq(x2, re(Piecewise((a - 3, a >= 0), (nan, True))) + i*im(Piecewise((a - 3, a >= 0), (nan, True))))
    © Sr Examen – Калькулятор