sqrt(4x-1)=y la ecuación
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
2 2
1 im (y) re (y) I*im(y)*re(y)
x1 = - - ------ + ------ + -------------
4 4 4 2
$$x_{1} = \frac{\left(\operatorname{re}{\left(y\right)}\right)^{2}}{4} + \frac{i \operatorname{re}{\left(y\right)} \operatorname{im}{\left(y\right)}}{2} - \frac{\left(\operatorname{im}{\left(y\right)}\right)^{2}}{4} + \frac{1}{4}$$
x1 = re(y)^2/4 + i*re(y)*im(y)/2 - im(y)^2/4 + 1/4
Suma y producto de raíces
[src]
2 2
1 im (y) re (y) I*im(y)*re(y)
- - ------ + ------ + -------------
4 4 4 2
$$\frac{\left(\operatorname{re}{\left(y\right)}\right)^{2}}{4} + \frac{i \operatorname{re}{\left(y\right)} \operatorname{im}{\left(y\right)}}{2} - \frac{\left(\operatorname{im}{\left(y\right)}\right)^{2}}{4} + \frac{1}{4}$$
2 2
1 im (y) re (y) I*im(y)*re(y)
- - ------ + ------ + -------------
4 4 4 2
$$\frac{\left(\operatorname{re}{\left(y\right)}\right)^{2}}{4} + \frac{i \operatorname{re}{\left(y\right)} \operatorname{im}{\left(y\right)}}{2} - \frac{\left(\operatorname{im}{\left(y\right)}\right)^{2}}{4} + \frac{1}{4}$$
2 2
1 im (y) re (y) I*im(y)*re(y)
- - ------ + ------ + -------------
4 4 4 2
$$\frac{\left(\operatorname{re}{\left(y\right)}\right)^{2}}{4} + \frac{i \operatorname{re}{\left(y\right)} \operatorname{im}{\left(y\right)}}{2} - \frac{\left(\operatorname{im}{\left(y\right)}\right)^{2}}{4} + \frac{1}{4}$$
2 2
1 im (y) re (y) I*im(y)*re(y)
- - ------ + ------ + -------------
4 4 4 2
$$\frac{\left(\operatorname{re}{\left(y\right)}\right)^{2}}{4} + \frac{i \operatorname{re}{\left(y\right)} \operatorname{im}{\left(y\right)}}{2} - \frac{\left(\operatorname{im}{\left(y\right)}\right)^{2}}{4} + \frac{1}{4}$$
1/4 - im(y)^2/4 + re(y)^2/4 + i*im(y)*re(y)/2