3202x + 1800 x2x^{2} + 400 x3x^{3} = -3192

asked by guest
on Jan 31, 2025 at 5:10 pm



You asked:

Solve the equation 3202x+1800x2+400x3=−31923202 x + 1800 {x}^{2} + 400 {x}^{3} = -3192 for the variable xx.

MathBot Answer:

The 3 solutions to the equation are: x=−32−29403800+32496369531400033+25120029403800+3249636953140003≈−2.6022484x=−32−25140029403800+3249636953140003+29403800+32496369531400036+i(251340029403800+3249636953140003+329403800+32496369531400036)≈−0.94887581+1.4718062ix=−32−25140029403800+3249636953140003+29403800+32496369531400036+i(−329403800+32496369531400036−251340029403800+3249636953140003)≈−0.94887581−1.4718062i\begin{aligned}x &= - \frac{3}{2} - \frac{\sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}}{3} + \frac{251}{200 \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}} \approx -2.6022484\\x &= - \frac{3}{2} - \frac{251}{400 \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}} + \frac{\sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}}{6} + i \left(\frac{251 \sqrt{3}}{400 \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}} + \frac{\sqrt{3} \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}}{6}\right) \approx -0.94887581 + 1.4718062 i\\x &= - \frac{3}{2} - \frac{251}{400 \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}} + \frac{\sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}}{6} + i \left(- \frac{\sqrt{3} \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}}{6} - \frac{251 \sqrt{3}}{400 \sqrt[3]{\frac{29403}{800} + \frac{3 \sqrt{2496369531}}{4000}}}\right) \approx -0.94887581 -1.4718062 i\end{aligned}

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

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