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Question Number 150191 by mathdanisur last updated on 10/Aug/21

{ ((x - (√y) = 7)),((y + (√x) = 7)) :} ⇒ xy = ?

$$\begin{cases}{\mathrm{x}\:-\:\sqrt{\mathrm{y}}\:=\:\mathrm{7}}\\{\mathrm{y}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 10/Aug/21

{ ((x - (√y)_(v) = 7)),((y + (√x)_(u) = 7)) :} ⇒ xy = ? u^2 −v=7 ∧ v^2 +u=7 u^2 −v^2 −v−u=0 (u−v)(u+v)−(u+v)=0 (u+v)(u−v−1)=0 u=−v ∨ u−v=1 ^• (u=−v∧u^2 −v=7) ∨^(••) u=v+1 v^2 −v−7=0 v=((1±(√(1+28)))/2)=((1±(√(29)))/2)=(√y) u=−((1±(√(29)))/2)=(√x) x=(((1+(√(29)))/2))^2 ,v=(((1+(√(29)))/2))^2 x=y=((1+29+2(√(29)))/4)=((15+(√(29)))/2) ^(••) u=v+1 ∧ v^2 +u=7 v^2 +v−6=0 (v+3)(v−2)=0 v=−3 ∨ v=2 (v=−3∧v^2 +u=7)∨(v=2∧v^2 +u=7) (−3)^2 +u=7 ∨ 2^2 +u=7 u=−2 ∨ u=3 (u,v)=(−2,−3),(3,2) ((√x),(√y))=(−2,−3)_(rejected) ∨ ((√x),(√y))=(2,3) (x,y)=(4,9),((((15+(√(29)))/2),((15+(√(29)))/2)) xy=4×9=36 xy=(((15+(√(29)))/2))^2 There′s possibility of extraneous roots.

$$\begin{cases}{\mathrm{x}\:-\:\underset{\mathrm{v}} {\underbrace{\sqrt{\mathrm{y}}}}\:=\:\mathrm{7}}\\{\mathrm{y}\:+\:\underset{\mathrm{u}} {\underbrace{\sqrt{\mathrm{x}}}}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:? \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{v}=\mathrm{7}\:\wedge\:\mathrm{v}^{\mathrm{2}} +\mathrm{u}=\mathrm{7} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} −\mathrm{v}−\mathrm{u}=\mathrm{0} \\ $$$$\left(\mathrm{u}−\mathrm{v}\right)\left(\mathrm{u}+\mathrm{v}\right)−\left(\mathrm{u}+\mathrm{v}\right)=\mathrm{0} \\ $$$$\left(\mathrm{u}+\mathrm{v}\right)\left(\mathrm{u}−\mathrm{v}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{u}=−\mathrm{v}\:\vee\:\mathrm{u}−\mathrm{v}=\mathrm{1} \\ $$$$\:^{\bullet} \left(\mathrm{u}=−\mathrm{v}\wedge\mathrm{u}^{\mathrm{2}} −\mathrm{v}=\mathrm{7}\right)\:\:\:\vee\:^{\bullet\bullet} \mathrm{u}=\mathrm{v}+\mathrm{1} \\ $$$$\:\:\:\:\mathrm{v}^{\mathrm{2}} −\mathrm{v}−\mathrm{7}=\mathrm{0}\: \\ $$$$\:\:\:\mathrm{v}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{28}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{29}}}{\mathrm{2}}=\sqrt{\mathrm{y}} \\ $$$$\:\:\:\mathrm{u}=−\frac{\mathrm{1}\pm\sqrt{\mathrm{29}}}{\mathrm{2}}=\sqrt{\mathrm{x}} \\ $$$$\:\:\:\:\mathrm{x}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)^{\mathrm{2}} ,\mathrm{v}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{x}=\mathrm{y}=\frac{\mathrm{1}+\mathrm{29}+\mathrm{2}\sqrt{\mathrm{29}}}{\mathrm{4}}=\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}} \\ $$$$\:\:^{\bullet\bullet} \mathrm{u}=\mathrm{v}+\mathrm{1}\:\wedge\:\mathrm{v}^{\mathrm{2}} +\mathrm{u}=\mathrm{7}\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{v}^{\mathrm{2}} +\mathrm{v}−\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left(\mathrm{v}+\mathrm{3}\right)\left(\mathrm{v}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{v}=−\mathrm{3}\:\vee\:\mathrm{v}=\mathrm{2}\:\:\:\:\: \\ $$$$\:\left(\mathrm{v}=−\mathrm{3}\wedge\mathrm{v}^{\mathrm{2}} +\mathrm{u}=\mathrm{7}\right)\vee\left(\mathrm{v}=\mathrm{2}\wedge\mathrm{v}^{\mathrm{2}} +\mathrm{u}=\mathrm{7}\right) \\ $$$$\:\:\:\:\left(−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{u}=\mathrm{7}\:\:\vee\:\mathrm{2}^{\mathrm{2}} +\mathrm{u}=\mathrm{7} \\ $$$$\:\:\:\:\mathrm{u}=−\mathrm{2}\:\:\vee\:\mathrm{u}=\mathrm{3} \\ $$$$\left(\mathrm{u},\mathrm{v}\right)=\left(−\mathrm{2},−\mathrm{3}\right),\left(\mathrm{3},\mathrm{2}\right) \\ $$$$\underset{\mathrm{rejected}} {\underbrace{\left(\sqrt{\mathrm{x}},\sqrt{\mathrm{y}}\right)=\left(−\mathrm{2},−\mathrm{3}\right)}}\:\vee\:\left(\sqrt{\mathrm{x}},\sqrt{\mathrm{y}}\right)=\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\:\:\:\:\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{4},\mathrm{9}\right),\left(\left(\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}},\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)\right. \\ $$$$\:\:\:\:\mathrm{xy}=\mathrm{4}×\mathrm{9}=\mathrm{36} \\ $$$$\:\:\:\:\mathrm{xy}=\left(\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathcal{T}{here}'{s}\:{possibility}\:{of}\:{extraneous}\: \\ $$$${roots}. \\ $$

Commented by mathdanisur last updated on 10/Aug/21

Thank You Ser, ans 36

$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{ans}\:\mathrm{36} \\ $$

Answered by MJS_new last updated on 10/Aug/21

x−(√y)=7 ⇒ y=(x−7)^2 ∧x≥7 ⇒ ⇒ x=((x−7)^2 −7)^2 x^4 −28x^3 +280x^2 −1177x+1764=0 (x−9)(x−4)(x^2 −15x+49)=0 ⇒ x=9[∨x=((15+(√(29)))/2) wrong because y≤7] ⇒ y=4[∨y=((15+(√(29)))/2)>7 ⇒ wrong] ⇒ x=9∧y=4 no other solutions xy=36

$${x}−\sqrt{{y}}=\mathrm{7}\:\Rightarrow\:{y}=\left({x}−\mathrm{7}\right)^{\mathrm{2}} \wedge{x}\geqslant\mathrm{7} \\ $$$$\Rightarrow \\ $$$$\Rightarrow \\ $$$${x}=\left(\left({x}−\mathrm{7}\right)^{\mathrm{2}} −\mathrm{7}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{28}{x}^{\mathrm{3}} +\mathrm{280}{x}^{\mathrm{2}} −\mathrm{1177}{x}+\mathrm{1764}=\mathrm{0} \\ $$$$\left({x}−\mathrm{9}\right)\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} −\mathrm{15}{x}+\mathrm{49}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{9}\left[\vee{x}=\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}}\:\mathrm{wrong}\:\mathrm{because}\:{y}\leqslant\mathrm{7}\right] \\ $$$$\Rightarrow\:{y}=\mathrm{4}\left[\vee{y}=\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}}>\mathrm{7}\:\Rightarrow\:\mathrm{wrong}\right] \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{9}\wedge{y}=\mathrm{4} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solutions} \\ $$$${xy}=\mathrm{36} \\ $$

Commented by mathdanisur last updated on 10/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$

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