Question-214419

Question Number 214419 by 2universe456 last updated on 08/Dec/24

Answered by golsendro last updated on 08/Dec/24

{ ((x^2 +y^2 +2xy=25)),((x^2 +y^2 +47=10xy)) :}⇒2xy+47=10xy+25 8xy= 22⇒4xy=11 4x(5−x)=11 ⇒4x^2 −20x+11=0 x_1 =(1/2)(5+(√(14)) ) ; x_2 =(1/2)(5−(√(14)) ) y_1 =5−(5/2)−(1/2)(√(14)) =(1/2)(5−(√(14)) ) y_2 = (1/2)(5+(√(14)) ) T= xy= (1/4)(25−14)= ((11)/4) H= x+y = 5 V= 2 ⇒V+T+H = 7+((11)/4) = ((39)/4)

$$\:\:\:\:\begin{cases}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}=\mathrm{25}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{47}=\mathrm{10xy}}\end{cases}\Rightarrow\mathrm{2xy}+\mathrm{47}=\mathrm{10xy}+\mathrm{25} \\ $$$$\:\:\:\mathrm{8xy}=\:\mathrm{22}\Rightarrow\mathrm{4xy}=\mathrm{11} \\ $$$$\:\:\:\mathrm{4x}\left(\mathrm{5}−\mathrm{x}\right)=\mathrm{11}\:\Rightarrow\mathrm{4x}^{\mathrm{2}} −\mathrm{20x}+\mathrm{11}=\mathrm{0} \\ $$$$\:\:\:\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}+\sqrt{\mathrm{14}}\:\right)\:;\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\sqrt{\mathrm{14}}\:\right) \\ $$$$\:\:\:\mathrm{y}_{\mathrm{1}} =\mathrm{5}−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{14}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\sqrt{\mathrm{14}}\:\right) \\ $$$$\:\:\:\mathrm{y}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}+\sqrt{\mathrm{14}}\:\right) \\ $$$$\:\:\mathrm{T}=\:\mathrm{xy}=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{25}−\mathrm{14}\right)=\:\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$\:\:\mathrm{H}=\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{5}\: \\ $$$$\:\:\:\mathrm{V}=\:\mathrm{2}\:\Rightarrow\mathrm{V}+\mathrm{T}+\mathrm{H}\:=\:\mathrm{7}+\frac{\mathrm{11}}{\mathrm{4}}\:=\:\frac{\mathrm{39}}{\mathrm{4}} \\ $$

Commented by Abdullahrussell last updated on 09/Dec/24

$$\:\:\:\:{xy}=\mathrm{6} \\ $$

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