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Question Number 12023 by tawa last updated on 09/Apr/17

Solve simultaneously  x + y − (√(xy)) = 3     .......... (i)  (√(x + 1)) + (√(y + 1)) = 4     .......... (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:−\:\sqrt{\mathrm{xy}}\:=\:\mathrm{3}\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$$$\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:+\:\sqrt{\mathrm{y}\:+\:\mathrm{1}}\:=\:\mathrm{4}\:\:\:\:\:..........\:\left(\mathrm{ii}\right) \\ $$

Answered by Mr Chheang Chantria last updated on 10/Apr/17

(i). x+y−(√(xy))≥x+y−((x+y)/2)=((x+y)/2)         3             ≥ ((x+y)/2)  ⇔x+y≤6    . (√(x+1))+(√(y+1)) ≤(√(2(x+y+2)))                                  ≤(√(2(6+2))) = 4  ⇒(√(x+1))+(√(y+1))≤4                                       by (ii) the sign equal when x=y=3

$$\left(\mathrm{i}\right).\:\boldsymbol{{x}}+\boldsymbol{{y}}−\sqrt{\boldsymbol{{xy}}}\geqslant\boldsymbol{{x}}+\boldsymbol{{y}}−\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}=\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\:\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}} \\ $$$$\Leftrightarrow\boldsymbol{{x}}+\boldsymbol{{y}}\leqslant\mathrm{6}\:\: \\ $$$$.\:\sqrt{\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{y}}+\mathrm{1}}\:\leqslant\sqrt{\mathrm{2}\left(\boldsymbol{{x}}+\boldsymbol{{y}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\sqrt{\mathrm{2}\left(\mathrm{6}+\mathrm{2}\right)}\:=\:\mathrm{4} \\ $$$$\Rightarrow\sqrt{\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{y}}+\mathrm{1}}\leqslant\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{{by}}\:\left(\boldsymbol{{ii}}\right)\:\boldsymbol{{the}}\:\boldsymbol{{sign}}\:\boldsymbol{{equal}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\boldsymbol{{y}}=\mathrm{3} \\ $$

Commented by Mr Chheang Chantria last updated on 10/Apr/17

this is for student.   we have ((√x)−(√y))^2 ≥0   for x,y≥0                   x−2(√(xy))+y≥0                      x+y≥2(√(xy))         or ((x+y)/2)≥(√(xy))         or −((x+y)/2)≤−(√(xy))

$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{for}}\:\boldsymbol{{student}}. \\ $$$$\:\boldsymbol{{we}}\:\boldsymbol{{have}}\:\left(\sqrt{\boldsymbol{{x}}}−\sqrt{\boldsymbol{{y}}}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:\:\mathrm{for}\:\boldsymbol{{x}},\boldsymbol{{y}}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}−\mathrm{2}\sqrt{\boldsymbol{{xy}}}+\boldsymbol{{y}}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}\geqslant\mathrm{2}\sqrt{\boldsymbol{{xy}}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{or}}\:\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}\geqslant\sqrt{\boldsymbol{{xy}}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{or}}\:−\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}\leqslant−\sqrt{\boldsymbol{{xy}}} \\ $$$$ \\ $$

Commented by tawa last updated on 10/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by ajfour last updated on 09/Apr/17

squaring ...(ii)  x+y+2+2(√((x+2)(y+1))) =16  3+(√(xy))+2+2(√((x+1)(y+1))) =16  2(√((x+1)(y+1))) =11−(√(xy))  squaring :  4(xy+1+x+y)=121+xy−22(√(xy))  4(xy+1+3+(√(xy)) =121+xy−22(√(xy))  3xy+26(√(xy))−105=0  if (√(xy)) =t  3t^2 +26t−105=0  t=(√(xy))=((−26+(√(676+1260)))/6)  (√(xy)) =((−26+44)/6)=3     ....(iii)  xy=9  from ....(i)  x+y=3+(√(xy))  ⇒ x+y=6   ....(iv)  (x−y)^2 =(x+y)^2 −4xy  so, (x−y)^2 =36−4(9)=0  ⇒  x=y  and as  x+y=6     x=y=3  .

$${squaring}\:...\left({ii}\right) \\ $$$${x}+{y}+\mathrm{2}+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\right)\left({y}+\mathrm{1}\right)}\:=\mathrm{16} \\ $$$$\mathrm{3}+\sqrt{{xy}}+\mathrm{2}+\mathrm{2}\sqrt{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)}\:=\mathrm{16} \\ $$$$\mathrm{2}\sqrt{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)}\:=\mathrm{11}−\sqrt{{xy}} \\ $$$${squaring}\:: \\ $$$$\mathrm{4}\left({xy}+\mathrm{1}+{x}+{y}\right)=\mathrm{121}+{xy}−\mathrm{22}\sqrt{{xy}} \\ $$$$\mathrm{4}\left({xy}+\mathrm{1}+\mathrm{3}+\sqrt{{xy}}\:=\mathrm{121}+{xy}−\mathrm{22}\sqrt{{xy}}\right. \\ $$$$\mathrm{3}{xy}+\mathrm{26}\sqrt{{xy}}−\mathrm{105}=\mathrm{0} \\ $$$${if}\:\sqrt{{xy}}\:={t} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} +\mathrm{26}{t}−\mathrm{105}=\mathrm{0} \\ $$$${t}=\sqrt{{xy}}=\frac{−\mathrm{26}+\sqrt{\mathrm{676}+\mathrm{1260}}}{\mathrm{6}} \\ $$$$\sqrt{{xy}}\:=\frac{−\mathrm{26}+\mathrm{44}}{\mathrm{6}}=\mathrm{3}\:\:\:\:\:....\left({iii}\right) \\ $$$${xy}=\mathrm{9} \\ $$$${from}\:....\left({i}\right) \\ $$$${x}+{y}=\mathrm{3}+\sqrt{{xy}} \\ $$$$\Rightarrow\:{x}+{y}=\mathrm{6}\:\:\:....\left({iv}\right) \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy} \\ $$$${so},\:\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{36}−\mathrm{4}\left(\mathrm{9}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}={y} \\ $$$${and}\:{as}\:\:{x}+{y}=\mathrm{6} \\ $$$$\:\:\:{x}={y}=\mathrm{3}\:\:. \\ $$

Commented by tawa last updated on 09/Apr/17

wow. God bless you sir.

$$\mathrm{wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Apr/17

x+1+y+1+2(√(x+1))(√(y+1))=16  (√(xy))+5+2(√(x+1))(√(y+1))=16  2(√(x+1))(√(y+1))=11−(√(xy))  4(xy+x+y+1)=121−22(√(xy))+xy  x+y=t  4[(t−3)^2 +t+1]=121−22(t−3)+(t−3)^2   3(t−3)^2 +26(t−3)−105=0  t−3=((−26±(√(26^2 +4×3×105)))/(2×3))=((−26±44)/6)=3,−11.66  t=x+y=6⇒(√(xy))=3⇒xy=9  ⇒x=y=3   ■

$${x}+\mathrm{1}+{y}+\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{y}+\mathrm{1}}=\mathrm{16} \\ $$$$\sqrt{{xy}}+\mathrm{5}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{y}+\mathrm{1}}=\mathrm{16} \\ $$$$\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{y}+\mathrm{1}}=\mathrm{11}−\sqrt{{xy}} \\ $$$$\mathrm{4}\left({xy}+{x}+{y}+\mathrm{1}\right)=\mathrm{121}−\mathrm{22}\sqrt{{xy}}+{xy} \\ $$$${x}+{y}={t} \\ $$$$\mathrm{4}\left[\left({t}−\mathrm{3}\right)^{\mathrm{2}} +{t}+\mathrm{1}\right]=\mathrm{121}−\mathrm{22}\left({t}−\mathrm{3}\right)+\left({t}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}\left({t}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{26}\left({t}−\mathrm{3}\right)−\mathrm{105}=\mathrm{0} \\ $$$${t}−\mathrm{3}=\frac{−\mathrm{26}\pm\sqrt{\mathrm{26}^{\mathrm{2}} +\mathrm{4}×\mathrm{3}×\mathrm{105}}}{\mathrm{2}×\mathrm{3}}=\frac{−\mathrm{26}\pm\mathrm{44}}{\mathrm{6}}=\mathrm{3},−\mathrm{11}.\mathrm{66} \\ $$$${t}={x}+{y}=\mathrm{6}\Rightarrow\sqrt{{xy}}=\mathrm{3}\Rightarrow{xy}=\mathrm{9} \\ $$$$\Rightarrow{x}={y}=\mathrm{3}\:\:\:\blacksquare \\ $$$$ \\ $$$$ \\ $$

Commented by tawa last updated on 09/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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