1.5x −2(x−50.0)^3−74.5 = 0.75

asked by guest
on Mar 16, 2025 at 7:33 am



You asked:

Solve the equation 1.5x2(x50.0)374.5=0.751.5 x - 2 {\left( x - 50.0 \right)}^{3} - 74.5 = 0.75 for the variable xx.

MathBot Answer:

The 3 solutions to the equation are: x=50cos(π9)49.060307x=3sin(π9)43re(1(123i2)2716+273i163)4+cos(π9)4+50+i(3im(1(123i2)2716+273i163)4+sin(π9)4+3cos(π9)4)50.1736482.010143ix=3sin(π9)4+cos(π9)43re(1(12+3i2)2716+273i163)4+50+i(3cos(π9)4+sin(π9)43im(1(12+3i2)2716+273i163)4)50.766044+2.010142i\begin{aligned}x &= 50 - \cos{\left(\frac{\pi}{9} \right)} \approx 49.060307\\x &= - \frac{\sqrt{3} \sin{\left(\frac{\pi}{9} \right)}}{4} - \frac{3 \operatorname{re}{\left(\frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{16} + \frac{27 \sqrt{3} i}{16}}}\right)}}{4} + \frac{\cos{\left(\frac{\pi}{9} \right)}}{4} + 50 + i \left(- \frac{3 \operatorname{im}{\left(\frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{16} + \frac{27 \sqrt{3} i}{16}}}\right)}}{4} + \frac{\sin{\left(\frac{\pi}{9} \right)}}{4} + \frac{\sqrt{3} \cos{\left(\frac{\pi}{9} \right)}}{4}\right) \approx 50.173648 -2.0 \cdot 10^{-143} i\\x &= \frac{\sqrt{3} \sin{\left(\frac{\pi}{9} \right)}}{4} + \frac{\cos{\left(\frac{\pi}{9} \right)}}{4} - \frac{3 \operatorname{re}{\left(\frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{16} + \frac{27 \sqrt{3} i}{16}}}\right)}}{4} + 50 + i \left(- \frac{\sqrt{3} \cos{\left(\frac{\pi}{9} \right)}}{4} + \frac{\sin{\left(\frac{\pi}{9} \right)}}{4} - \frac{3 \operatorname{im}{\left(\frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{16} + \frac{27 \sqrt{3} i}{16}}}\right)}}{4}\right) \approx 50.766044 + 2.0 \cdot 10^{-142} i\end{aligned}


ii is the imaginary unit, defined as i2=1i^2 = -1.