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Question Number 202873 by lorance last updated on 05/Jan/24

$$if{x}^4-x-3=0$$$$\mathrm{S}\mathrm{\&}\mathrm{P}are+\mathrm{\&}\times realroots$$$$whatisvalueof4P^2+S^2?$$

Commented by Frix last updated on 05/Jan/24

$$x^4+px+q=0$$$$x_1=a$$$$x_2=b$$$$x_{3,4}=-\frac{a+b}{2}\pm \frac{\sqrt{3a^2+2ab+3b^2}}{2}\mathrm{i}$$$$\Rightarrow$$$$x^4-(a+b)(a^2+b^2)x+a(a^2+ab+b^2)b=0$$$$p=-(a+b)(a^2+b^2)\wedge q=a(a^2+ab+b^2)b$$$$\mathrm{You}\mathrm{ask}\mathrm{for}4a^2{b}^2+{(a+b)}^2=r$$$$\mathrm{Let}u=a+b\wedge v=ab$$$$p=2uv-u^3$$$$q=u^{2}v-v^2$$$$\mathrm{This}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{lead}\mathrm{to}\mathrm{any}\mathrm{nice}\mathrm{solution}\mathrm{for}$$$$r=u^2+4v^2\mathrm{generally}\mathrm{and}\mathrm{also}\mathrm{not}\mathrm{with}\mathrm{the}$$$$\mathrm{given}p=-1\wedge q=-3:$$$$2uv-u^3=-1\Rightarrow v=\frac{u^3-1}{2u}$$$$u^{2}v-v^2=-3\Rightarrow u^6+12u^3-1=0\Rightarrow$$$$u=\pm \sqrt{\sqrt[3]{\frac{\sqrt{257}}{2}+\frac{1}{2}}-\sqrt[3]{\frac{\sqrt{257}}{2}-\frac{1}{2}}}$$$$r=u^4+u^2-2u+\frac{1}{u^2}$$$$\mathrm{No}\mathrm{chance}\mathrm{for}\mathrm{something}\mathrm{useable}...$$

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