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Question Number 106281 by bemath last updated on 04/Aug/20

(√(x+50))+(√(y+100))+(√(z+150))=((x+y+z)/4)+78 find x+y−z

$$\sqrt{\mathrm{x}+\mathrm{50}}+\sqrt{\mathrm{y}+\mathrm{100}}+\sqrt{\mathrm{z}+\mathrm{150}}=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{4}}+\mathrm{78} \\ $$$$\mathrm{find}\:\mathrm{x}+\mathrm{y}−\mathrm{z}\: \\ $$

Answered by bobhans last updated on 04/Aug/20

4(√(x+50))+4(√(y+100))+4(√(z+150)) =x+y+z+312 (x+50)−4(√(x+50)))+(y+100)−4(√(y+100)))+z+150−4(√(z+150))+12=0 set { ((x+50=a^2 ;),(y+100=b^2 ;),(z+150=c^2 })) :} (a^2 −4a+4)+(b^2 −4b+4)+(c^2 −4c+4)=0 (a−2)^2 +(b−2)^2 +(c−2)^2 =0 a = 2 →x+50=4 ; x=−46 b=2 →y+100=4; y=−96 c=2 →z+150=4; z=−146 therefore x+y−z=−142+146=4.★

$$\mathrm{4}\sqrt{\mathrm{x}+\mathrm{50}}+\mathrm{4}\sqrt{\mathrm{y}+\mathrm{100}}+\mathrm{4}\sqrt{\mathrm{z}+\mathrm{150}}\:=\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{312} \\ $$$$\left.\left(\left.\mathrm{x}+\mathrm{50}\right)−\mathrm{4}\sqrt{\mathrm{x}+\mathrm{50}}\right)+\left(\mathrm{y}+\mathrm{100}\right)−\mathrm{4}\sqrt{\mathrm{y}+\mathrm{100}}\right)+\mathrm{z}+\mathrm{150}−\mathrm{4}\sqrt{\mathrm{z}+\mathrm{150}}+\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{set}\:\begin{cases}{\mathrm{x}+\mathrm{50}=\mathrm{a}^{\mathrm{2}} ;}&{\mathrm{y}+\mathrm{100}=\mathrm{b}^{\mathrm{2}} ;}&{\left.\mathrm{z}+\mathrm{150}=\mathrm{c}^{\mathrm{2}} \:\right\}}\end{cases} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} −\mathrm{4a}+\mathrm{4}\right)+\left(\mathrm{b}^{\mathrm{2}} −\mathrm{4b}+\mathrm{4}\right)+\left(\mathrm{c}^{\mathrm{2}} −\mathrm{4c}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\left(\mathrm{a}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{c}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{a}\:=\:\mathrm{2}\:\rightarrow\mathrm{x}+\mathrm{50}=\mathrm{4}\:;\:\mathrm{x}=−\mathrm{46} \\ $$$$\mathrm{b}=\mathrm{2}\:\rightarrow\mathrm{y}+\mathrm{100}=\mathrm{4};\:\mathrm{y}=−\mathrm{96} \\ $$$$\mathrm{c}=\mathrm{2}\:\rightarrow\mathrm{z}+\mathrm{150}=\mathrm{4};\:\mathrm{z}=−\mathrm{146} \\ $$$$\mathrm{therefore}\:\mathrm{x}+\mathrm{y}−\mathrm{z}=−\mathrm{142}+\mathrm{146}=\mathrm{4}.\bigstar \\ $$

Commented by bemath last updated on 04/Aug/20

waw....super..

$$\mathrm{waw}....\mathrm{super}.. \\ $$

Answered by john santu last updated on 04/Aug/20

another way set (√(x+50)) = p ; (√(y+100 )) = q and (√(z+150)) = r then ⇒ p+q+r = ((p^2 +q^2 +r^2 −300)/4)+78 p^2 +q^2 +r^2 −4p−4q−4r+12=0 (p^2 −4p+4)+(q^2 −4q+12)+(r^2 −4r+4)=0 (p−2)^2 +(q−2)^2 +(r−2)^2 = 0 { ((p=2 ; q=2 ; r=2 })) :} x = −46 ; y = −96 & z = −146 x+y−z = −142+146 = 4

$$\mathrm{another}\:\mathrm{way}\: \\ $$$$\mathrm{set}\:\sqrt{\mathrm{x}+\mathrm{50}}\:=\:\mathrm{p}\:;\:\sqrt{\mathrm{y}+\mathrm{100}\:}\:=\:\mathrm{q} \\ $$$$\mathrm{and}\:\sqrt{\mathrm{z}+\mathrm{150}}\:=\:\mathrm{r}\: \\ $$$$\mathrm{then}\:\Rightarrow\:\mathrm{p}+\mathrm{q}+\mathrm{r}\:=\:\frac{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} −\mathrm{300}}{\mathrm{4}}+\mathrm{78} \\ $$$$\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} −\mathrm{4p}−\mathrm{4q}−\mathrm{4r}+\mathrm{12}=\mathrm{0} \\ $$$$\left(\mathrm{p}^{\mathrm{2}} −\mathrm{4p}+\mathrm{4}\right)+\left(\mathrm{q}^{\mathrm{2}} −\mathrm{4q}+\mathrm{12}\right)+\left(\mathrm{r}^{\mathrm{2}} −\mathrm{4r}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\left(\mathrm{p}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{q}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{r}−\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\begin{cases}{\left.\mathrm{p}=\mathrm{2}\:;\:\mathrm{q}=\mathrm{2}\:;\:\mathrm{r}=\mathrm{2}\:\right\}}\end{cases} \\ $$$$\mathrm{x}\:=\:−\mathrm{46}\:;\:\mathrm{y}\:=\:−\mathrm{96}\:\&\:\mathrm{z}\:=\:−\mathrm{146}\: \\ $$$$\mathrm{x}+\mathrm{y}−\mathrm{z}\:=\:−\mathrm{142}+\mathrm{146}\:=\:\mathrm{4}\: \\ $$

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