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Question Number 55195 by behi83417@gmail.com last updated on 19/Feb/19

$${\mathbf{x}}^2+\mathbf{ax}+\sqrt{2}\mathbf{b}=0,\mathbf{has}2\mathbf{roots}:\mathbf{c}\mathbf{and}\mathbf{d},\mathbf{also}$$$${\mathbf{x}}^2+\mathbf{cx}+\sqrt{2}\mathbf{d}=0,\mathbf{has}2\mathbf{roots}:\mathbf{a}\mathbf{and}\mathbf{b}.\mathbf{such}$$$$\mathbf{that}:\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d},\mathbf{are}\mathbf{defferent}\mathbf{non}\mathbf{zero}$$$$\mathbf{numbers}.$$$$\mathbf{find}\mathbf{possible}\mathbf{value}\left(\mathbf{s}\right)\mathbf{for}:{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2+{\mathbf{d}}^2.$$

Answered by behi83417@gmail.com last updated on 19/Feb/19

$$c+d=-a,cd=\sqrt{2}b$$$$a+b=-c,ab=\sqrt{2}d$$$$abcd=2bd\Rightarrow ac=2\Rightarrow c=\frac{2}{a}$$$$a+b=-c\Rightarrow b=-\frac{2}{a}-a=-\frac{2+a^2}{a}$$$$ab=\sqrt{2}d\Rightarrow d=\frac{a.\frac{-(2+a^2)}{a}}{\sqrt{2}}=-\frac{2+a^2}{\sqrt{2}}$$$$c+d=-a\Rightarrow \frac{2}{a}-\frac{2+a^2}{\sqrt{2}}=-a$$$$2\sqrt{2}-a(2+a^2)=-a^2\sqrt{2}$$$$\Rightarrow a^3-\sqrt{2}{a}^2+2a-2\sqrt{2}=0$$$$\Rightarrow (a^2+2)(a-\sqrt{2})=0\Rightarrow a=\sqrt{2},a=\pm i\sqrt{2}$$$$d=-\frac{2+a^2}{\sqrt{2}}=-\frac{2+{(\sqrt{2}or\pm i\sqrt{2})}^2}{\sqrt{2}}=-2\sqrt{2},0$$$$b=-\frac{2+a^2}{a}=-2\sqrt{2},0,c=\frac{2}{a}=\sqrt{2},\mp i\sqrt{2}$$$$\Rightarrow a^2+b^2+c^2+d^2=20\vee -4.$$

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