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Question Number 79127 by ~blr237~ last updated on 22/Jan/20

$$Solveon\mathbb{R}\ast \mathbb{R}thefollowingsystem$$$$\{{}_{9^A+9^B+9^C=1}^{3^A+3^B+3^C=\sqrt{3}}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{because}\mathrm{x}>0,\mathrm{y}>0\mathrm{and}\mathrm{z}>0$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{x}+\mathrm{y}+\mathrm{z}=\sqrt{3}(\mathrm{xy}+\mathrm{xz}+\mathrm{yz})$$$$\mathrm{x}(\mathrm{y}\sqrt{3}-1)+\mathrm{y}(\mathrm{z}\sqrt{3}-1)+\mathrm{z}(\mathrm{x}\sqrt{3}-1)=0$$$$\Rightarrow \mathrm{y}\sqrt{3}-1=0$$$$\mathrm{z}\sqrt{3}-1=0$$$$\mathrm{x}\sqrt{3}-1=0$$

Commented by jagoll last updated on 23/Jan/20

$$9^{\mathrm{a}}+9^{\mathrm{b}}+9^{\mathrm{c}}={(3^{\mathrm{a}}+3^{\mathrm{b}}+3^{\mathrm{c}})}^2-2(3^{\mathrm{a}+\mathrm{b}}+3^{\mathrm{a}+\mathrm{c}}+3^{\mathrm{b}+\mathrm{c}})$$$$1=3-2(3^{\mathrm{a}+\mathrm{b}}+3^{\mathrm{a}+\mathrm{c}}+3^{\mathrm{b}+\mathrm{c}})$$$$3^{\mathrm{a}+\mathrm{b}}+3^{\mathrm{a}+\mathrm{c}}+3^{\mathrm{b}+\mathrm{c}}=1$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{let}{3}^{\mathrm{a}}=\mathrm{x},3^{\mathrm{b}}=\mathrm{y},3^{\mathrm{c}}=\mathrm{z}$$$$\mathrm{xy}+\mathrm{xz}+\mathrm{yz}=1$$$$\mathrm{x}+\mathrm{y}+\mathrm{z}=\sqrt{3}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1$$

Commented by jagoll last updated on 23/Jan/20

$${\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2-\mathrm{yz}+{\mathrm{z}}^2-\mathrm{xz}=0$$$$\mathrm{x}(\mathrm{x}-\mathrm{y})+\mathrm{x}(\mathrm{y}-\mathrm{z})+\mathrm{z}(\mathrm{z}-\mathrm{x})=0$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{x}=\mathrm{y}=\mathrm{z}$$$${\mathrm{x}}^2+{\mathrm{x}}^2+{\mathrm{x}}^2=1\Rightarrow \mathrm{x}=\frac{1}{\sqrt{3}}$$$$\therefore \mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{1}{\sqrt{3}}$$

Commented by ~blr237~ last updated on 23/Jan/20

$$yourexplanationofx=y=zisnotclearsir$$$$causeyou{don}^{\prime }tknowifx(y-z)>0orsamefortherest$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{impossible}\mathrm{x}(\mathrm{y}-\mathrm{z})>0\mathrm{sir}.$$$$\mathrm{because}\mathrm{sum}\mathrm{the}\mathrm{three}\mathrm{terms}\mathrm{is}$$$$\mathrm{zero}$$

Commented by ~blr237~ last updated on 23/Jan/20

$$youshouldprovethatthethreeofthemarepositivebeforeconcluding$$

Commented by MJS last updated on 23/Jan/20

$$\mathrm{it}\mathrm{must}\mathrm{be}\mathbb{R}\times \mathbb{R}\times \mathbb{R}\mathrm{not}\mathbb{R}\times \mathbb{R}$$

Answered by ~blr237~ last updated on 23/Jan/20

$$letfindA={(3^A-\frac{1}{\sqrt{3}})}^2+{(3^B-\frac{1}{\sqrt{3}})}^2+{(3^C-\frac{1}{\sqrt{3}})}^2$$$$A=(9^A-\frac{2}{\sqrt{3}}{3}^A+\frac{1}{3})+(9^B-\frac{2}{\sqrt{3}}{3}^B+\frac{1}{3})+(9^C-\frac{2}{\sqrt{3}}{3}^C+\frac{1}{3})$$$$=9^A+9^B+9^C-\frac{2}{\sqrt{3}}(3^A+3^B+3^C)+1$$$$=1-2+1=0$$$$suchaseachtermofAispositivewegot$$$$3^A=3^B=3^C=\frac{1}{\sqrt{3}}$$$$FinalyA=B=C=-\frac{1}{2}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{what}\mathrm{wrong}\mathrm{my}\mathrm{answer}?$$

Commented by ~blr237~ last updated on 23/Jan/20

$$we{don}^{\prime }tknowify\sqrt{3}-1>0,evenz\sqrt{3}-1andy\sqrt{3}-1$$$$thepropositionyouwanttois$$$$whenU⩾0,V⩾0$$$$U+V=0\Rightarrow U=0andV=0$$$$butyoudidverifyiffirstlyU,V(there{it}^{\prime }sx(y\sqrt{3}-1),y(z\sqrt{3}-1),z(x\sqrt{3}-1))werepositive$$

Commented by MJS last updated on 23/Jan/20

$$A={(3^A-\frac{1}{\sqrt{3}})}^2+{(3^B-\frac{1}{\sqrt{3}})}^2+{(3^C-\frac{1}{\sqrt{3}})}^2$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{do}\mathrm{that},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{confusing}\mathrm{and}\mathrm{simply}\mathrm{wrong}$$

Answered by MJS last updated on 23/Jan/20

$$3^A=x,3^B=y,3^C=z;x>0\wedge y>0\wedge z>0$$$$\{\begin{array}{l}x+y+z=\sqrt{3}\\ x^2+y^2+z^2=1\end{array}$$$$\{\begin{array}{l}z=\sqrt{3}-(x+y)\\ x^2+xy+y^2-\sqrt{3}x-\sqrt{3}y+1=0\end{array}$$$$\mathrm{let}y=sx;x>0$$$$\{\begin{array}{l}z=\sqrt{3}-(s+1)x\\ s^2+\frac{x-\sqrt{3}}{x}s+\frac{x^2-\sqrt{3}x+1}{x^2}=0\end{array}$$$$\mathrm{let}s=t-\frac{x-\sqrt{3}}{2x}$$$$\{\begin{array}{l}z=\frac{\sqrt{3}}{2}-(t+\frac{1}{2})x\\ t^2=-\frac{{(3x-\sqrt{3})}^2}{12x^2}\end{array}$$$$\Rightarrow t=\pm \frac{\sqrt{3}x-1}{2x}\mathrm{i}$$$$\Rightarrow s=-\frac{x-\sqrt{3}}{2x}\pm \frac{\sqrt{3}x-1}{2x}\mathrm{i}$$$$\Rightarrow y=-\frac{x-\sqrt{3}}{2}\pm \frac{\sqrt{3}x-1}{2}\mathrm{i}$$$$\Rightarrow z=-\frac{x-\sqrt{3}}{2}\mp \frac{\sqrt{3}x-1}{2}\mathrm{i}$$$$\Rightarrow \mathrm{we}\mathrm{only}\mathrm{get}\mathrm{a}\mathrm{real}\mathrm{solution}\mathrm{if}$$$$\frac{\sqrt{3}x-1}{2}=0\Rightarrow x=\frac{\sqrt{3}}{3}$$$$\Rightarrow s=1\wedge t=0\wedge x=y=z=\frac{\sqrt{3}}{3}$$$$\Rightarrow A=B=C=-\frac{1}{2}$$

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