Question-66489

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Question Number 66489 by rajesh4661kumar@gmail.com last updated on 16/Aug/19

Answered by $@ty@m123 last updated on 16/Aug/19

$${x}=\mathrm{4} \\ $$$${y}=\mathrm{9} \\ $$

Commented by rajesh4661kumar@gmail.com last updated on 16/Aug/19

$${how} \\ $$

Answered by MJS last updated on 16/Aug/19

first, try a few values for x and y, both are square numbers (1) y=11−(√x) x=1^2 ⇒ y=10 x=2^2 ⇒ y=3^2 (2) 2^2 +3=7 solution because of (2) x must be smaller than 7, so no other integer solution is possible second, without squaring (because this would lead to false solutions) draw the functions y=11−(√x) x=7−(√y) these are 2 half parabolas which only have one real intersection third, are there complex solutions? no because the full parabolas have 4 real intersections there′s “no place” for other solutions

$$\mathrm{first},\:\mathrm{try}\:\mathrm{a}\:\mathrm{few}\:\mathrm{values}\:\mathrm{for}\:{x}\:\mathrm{and}\:{y},\:\mathrm{both}\:\mathrm{are} \\ $$$$\mathrm{square}\:\mathrm{numbers} \\ $$$$\left(\mathrm{1}\right)\:\:{y}=\mathrm{11}−\sqrt{{x}} \\ $$$$\:\:\:\:\:{x}=\mathrm{1}^{\mathrm{2}} \:\Rightarrow\:{y}=\mathrm{10} \\ $$$$\:\:\:\:\:{x}=\mathrm{2}^{\mathrm{2}} \:\Rightarrow\:{y}=\mathrm{3}^{\mathrm{2}} \:\:\left(\mathrm{2}\right)\:\mathrm{2}^{\mathrm{2}} +\mathrm{3}=\mathrm{7}\:\mathrm{solution} \\ $$$$\mathrm{because}\:\mathrm{of}\:\left(\mathrm{2}\right)\:{x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{smaller}\:\mathrm{than}\:\mathrm{7},\:\mathrm{so} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{integer}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{possible} \\ $$$$ \\ $$$$\mathrm{second},\:\mathrm{without}\:\mathrm{squaring}\:\left(\mathrm{because}\:\mathrm{this}\right. \\ $$$$\left.\mathrm{would}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{false}\:\mathrm{solutions}\right)\:\mathrm{draw}\:\mathrm{the} \\ $$$$\mathrm{functions} \\ $$$${y}=\mathrm{11}−\sqrt{{x}} \\ $$$${x}=\mathrm{7}−\sqrt{{y}} \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{2}\:\mathrm{half}\:\mathrm{parabolas}\:\mathrm{which}\:\mathrm{only}\:\mathrm{have} \\ $$$$\mathrm{one}\:\mathrm{real}\:\mathrm{intersection} \\ $$$$ \\ $$$$\mathrm{third},\:\mathrm{are}\:\mathrm{there}\:\mathrm{complex}\:\mathrm{solutions}? \\ $$$$\mathrm{no}\:\mathrm{because}\:\mathrm{the}\:\mathrm{full}\:\mathrm{parabolas}\:\mathrm{have}\:\mathrm{4}\:\mathrm{real} \\ $$$$\mathrm{intersections}\:\mathrm{there}'\mathrm{s}\:“\mathrm{no}\:\mathrm{place}''\:\mathrm{for}\:\mathrm{other} \\ $$$$\mathrm{solutions} \\ $$

Answered by behi83417@gmail.com last updated on 16/Aug/19

((√x)−2)+((√y)−3)((√y)+3)=0 ((√x)−2)((√x)+2)+((√y)−3)=0 ⇒−((√y)−3)((√y)+3)((√x)+2)+((√y)−3)=0 ⇒((√y)−3)[−((√y)+3)((√x)+2)+1]=0 ⇒(√y)−3=0⇒y=9 (√x)=11−9=2⇒x=4 .■

$$\left(\sqrt{\mathrm{x}}−\mathrm{2}\right)+\left(\sqrt{\mathrm{y}}−\mathrm{3}\right)\left(\sqrt{\mathrm{y}}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{x}}−\mathrm{2}\right)\left(\sqrt{\mathrm{x}}+\mathrm{2}\right)+\left(\sqrt{\mathrm{y}}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow−\left(\sqrt{\mathrm{y}}−\mathrm{3}\right)\left(\sqrt{\mathrm{y}}+\mathrm{3}\right)\left(\sqrt{\mathrm{x}}+\mathrm{2}\right)+\left(\sqrt{\mathrm{y}}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{y}}−\mathrm{3}\right)\left[−\left(\sqrt{\mathrm{y}}+\mathrm{3}\right)\left(\sqrt{\mathrm{x}}+\mathrm{2}\right)+\mathrm{1}\right]=\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{y}}−\mathrm{3}=\mathrm{0}\Rightarrow\mathrm{y}=\mathrm{9} \\ $$$$\sqrt{\mathrm{x}}=\mathrm{11}−\mathrm{9}=\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{4}\:\:.\blacksquare \\ $$

Commented by Prithwish sen last updated on 16/Aug/19

but sir x+(1/( (√x)+2)) −3 =7 ⇒x + (1/( (√x)+2)) = 10

$$\mathrm{but}\:\mathrm{sir}\: \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}+\mathrm{2}}\:−\mathrm{3}\:=\mathrm{7}\:\:\Rightarrow\mathrm{x}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}+\mathrm{2}}\:=\:\mathrm{10} \\ $$

Commented by behi83417@gmail.com last updated on 21/Aug/19

(√y)+3=(1/( (√x)+2))⇒x+(1/( (√x)+2))+3=7 ⇒x(√x)+2x=4(√x)+8⇒ (x−4)(√x)+2(x−4)=0⇒(x−4)((√x)+2)=0 ⇒^((√x)+2≠0) x−4=0⇒x=4 .■

$$\sqrt{\mathrm{y}}+\mathrm{3}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}+\mathrm{2}}\Rightarrow\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}+\mathrm{2}}+\mathrm{3}=\mathrm{7} \\ $$$$\Rightarrow\mathrm{x}\sqrt{\mathrm{x}}+\mathrm{2x}=\mathrm{4}\sqrt{\mathrm{x}}+\mathrm{8}\Rightarrow \\ $$$$\left(\mathrm{x}−\mathrm{4}\right)\sqrt{\mathrm{x}}+\mathrm{2}\left(\mathrm{x}−\mathrm{4}\right)=\mathrm{0}\Rightarrow\left(\mathrm{x}−\mathrm{4}\right)\left(\sqrt{\mathrm{x}}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\overset{\sqrt{\mathrm{x}}+\mathrm{2}\neq\mathrm{0}} {\Rightarrow}\:\:\:\:\:\mathrm{x}−\mathrm{4}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{4}\:\:\:\:\:.\blacksquare \\ $$

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