Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 70399 by Henri Boucatchou last updated on 04/Oct/19

$$\mathit{S}\mathit{o}\mathit{l}\mathit{v}\mathit{e}{\mathit{x}}^4+{\mathit{x}}^3-2{\mathit{a}\mathit{x}}^2-\mathit{a}\mathit{x}+{\mathit{a}}^2=0,\mathit{a}\in \mathbb{R}$$

Commented by Prithwish sen last updated on 04/Oct/19

$${\mathrm{x}}^4-{2\mathrm{ax}}^2+{\mathrm{a}}^2+{\mathrm{x}}^3-\mathrm{ax}=0$$$${({\mathrm{x}}^2-\mathrm{a})}^2+\mathrm{x}({\mathrm{x}}^2-\mathrm{a})=0$$$$({\mathrm{x}}^2-\mathrm{a})({\mathrm{x}}^2+\mathrm{x}-\mathrm{a})=0$$$$\Rightarrow \mathrm{x}=\pm \sqrt{}\mathrm{a}\mathrm{and}\mathrm{x}=\frac{-1\pm \sqrt{1+4\mathrm{a}}}{2}$$$$\mathbf{I}\mathbf{have}\mathbf{not}\mathbf{checked}\mathbf{the}\mathbf{answer}.\mathbf{Please}\mathbf{check}\mathbf{it}.$$

Commented by Prithwish sen last updated on 04/Oct/19

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}.\mathrm{I}\mathrm{correct}\mathrm{it}.$$

Commented by MJS last updated on 04/Oct/19

$$\mathrm{you}\mathrm{forgot}\mathrm{the}‘‘{\sqrt{}}^{″}$$

Answered by MJS last updated on 04/Oct/19

$$\mathrm{trying}\mathrm{factors}\mathrm{of}{a}^2$$$$x\in \pm \{a^2,a,\sqrt{a}\}$$$$\Rightarrow x=\pm \sqrt{a}$$$$\Rightarrow$$$$x^4+x^3-2{ax}^2-ax+a^2=$$$$=(x-\sqrt{a})(x+\sqrt{a})(x^2+x-a)=$$$$=(x-\sqrt{a})(x+\sqrt{a})(x+\frac{1}{2}-\frac{\sqrt{4a+1}}{2})(x+\frac{1}{2}+\frac{\sqrt{4a+1}}{2})$$$$\Rightarrow x=\pm \sqrt{a}\vee x=-\frac{1}{2}\pm \frac{\sqrt{4a+1}}{2}$$$$\Rightarrow \{\begin{array}{l}4\mathrm{real}\mathrm{solutions}\mathrm{for}0⩽a\\ 2\mathrm{real}\mathrm{and}2\mathrm{imaginary}\mathrm{solutions}\mathrm{for}-\frac{1}{4}⩽a<0\\ 2\mathrm{conjugated}\mathrm{complex}\mathrm{and}2\mathrm{imaginary}\mathrm{solutions}\mathrm{for}a<-\frac{1}{4}\end{array}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com