Question Number 80702 by john santu last updated on 05/Feb/20
$$\begin{cases}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\mathrm{34}}\\{\frac{\mathrm{1}}{\sqrt{{x}}}+\frac{\mathrm{1}}{\sqrt{{y}}}=\mathrm{23}−\frac{\mathrm{1}}{\sqrt{{xy}}}\:}\end{cases} \\ $$$${find}\:{the}\:{solution}. \\ $$
Commented by mind is power last updated on 05/Feb/20
$${let}\:\frac{\mathrm{1}}{\sqrt{{x}}}+\frac{\mathrm{1}}{\sqrt{{y}}}={u} \\ $$$$\frac{\mathrm{1}}{\sqrt{{xy}}}. \\ $$$$\Rightarrow{u}^{\mathrm{2}} −\mathrm{2}{v}=\mathrm{34} \\ $$$${u}=\mathrm{23}−{v} \\ $$$$\Rightarrow\left(\mathrm{23}−{v}\right)^{\mathrm{2}} −\mathrm{2}{v}=\mathrm{34} \\ $$$${v}^{\mathrm{2}} −\mathrm{48}{v}+\mathrm{495}=\mathrm{0} \\ $$$${v}\in\left\{\mathrm{15},\mathrm{28}\right\} \\ $$$${u}\geqslant\mathrm{0}\Rightarrow{v}=\mathrm{15}\:\Rightarrow{u}=\mathrm{8} \\ $$$$\begin{cases}{\frac{\mathrm{1}}{\sqrt{{x}}}+\frac{\mathrm{1}}{\sqrt{{y}}}=\mathrm{8}}\\{\frac{\mathrm{1}}{\sqrt{{x}}}.\frac{\mathrm{1}}{\sqrt{{y}}}=\mathrm{15}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\sqrt{{x}}},\frac{\mathrm{1}}{\sqrt{{y}}}\:\:{root}\:{of}\:\:{X}^{\mathrm{2}} −\mathrm{8}{X}+\mathrm{15} \\ $$$${X}\in\left\{\mathrm{3},\mathrm{5}\right\} \\ $$$$\Rightarrow\left({x},{y}\right)\in\left\{\left(\frac{\mathrm{1}}{\mathrm{9}},\frac{\mathrm{1}}{\mathrm{25}}\right);\left(\frac{\mathrm{1}}{\mathrm{25}},\frac{\mathrm{1}}{\mathrm{9}}\right)\right\} \\ $$$$ \\ $$
Commented by john santu last updated on 05/Feb/20
$${let}\:\frac{\mathrm{1}}{\sqrt{{x}}}+\frac{\mathrm{1}}{\sqrt{{y}}}={u},\:\frac{\mathrm{1}}{\sqrt{{xy}}}={v} \\ $$$${u}^{\mathrm{2}} −\mathrm{2}{v}=\mathrm{34}\:,\:{u}+{v}=\mathrm{23} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{80}=\mathrm{0}\:\Rightarrow\begin{cases}{{u}=\mathrm{8}\:}\\{{u}=−\mathrm{10}}\end{cases} \\ $$$${for}\:{u}\:=\:\mathrm{8}\:,\:{v}=\mathrm{15} \\ $$$$\frac{\mathrm{1}}{\sqrt{{x}}}+\frac{\mathrm{1}}{\sqrt{{y}}}=\mathrm{8}\:\wedge\frac{\mathrm{1}}{\sqrt{{xy}}}=\mathrm{15}\: \\ $$$$\frac{\mathrm{1}}{\sqrt{{x}}}=\mathrm{3}\:,\:\frac{\mathrm{1}}{\sqrt{{x}}}\:=\mathrm{5}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\:,\:{x}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\therefore\:\left(\frac{\mathrm{1}}{\mathrm{9}},\frac{\mathrm{1}}{\mathrm{25}}\right)\:,\:\left(\frac{\mathrm{1}}{\mathrm{25}},\:\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$
Commented by john santu last updated on 05/Feb/20
$${great}\:{sir} \\ $$
Commented by MJS last updated on 06/Feb/20
![(1/(√x))=u−v; (1/(√y))=u+v; u−v>0∧u+v>0 { ((v^2 =17−u^2 )),((u^2 +u−20=0)) :} { ((v_1 =±2(√2)i; v_2 =±1)),((u_1 =−5; u_2 =4)) :} [u_1 , v_1 not useable] (1/(√x))=5∧(1/(√y))=3 ∨ (1/(√x))=3∧(1/(√y))=5 x=(1/(25))∧y=(1/9) ∨ x=(1/9)∧y=(1/(25))](Q80755.png)
$$\frac{\mathrm{1}}{\sqrt{{x}}}={u}−{v};\:\frac{\mathrm{1}}{\sqrt{{y}}}={u}+{v};\:{u}−{v}>\mathrm{0}\wedge{u}+{v}>\mathrm{0} \\ $$$$\begin{cases}{{v}^{\mathrm{2}} =\mathrm{17}−{u}^{\mathrm{2}} }\\{{u}^{\mathrm{2}} +{u}−\mathrm{20}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{v}_{\mathrm{1}} =\pm\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i};\:{v}_{\mathrm{2}} =\pm\mathrm{1}}\\{{u}_{\mathrm{1}} =−\mathrm{5};\:{u}_{\mathrm{2}} =\mathrm{4}}\end{cases}\:\:\left[{u}_{\mathrm{1}} ,\:{v}_{\mathrm{1}} \:\mathrm{not}\:\mathrm{useable}\right] \\ $$$$\frac{\mathrm{1}}{\sqrt{{x}}}=\mathrm{5}\wedge\frac{\mathrm{1}}{\sqrt{{y}}}=\mathrm{3}\:\vee\:\frac{\mathrm{1}}{\sqrt{{x}}}=\mathrm{3}\wedge\frac{\mathrm{1}}{\sqrt{{y}}}=\mathrm{5} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{25}}\wedge{y}=\frac{\mathrm{1}}{\mathrm{9}}\:\vee\:{x}=\frac{\mathrm{1}}{\mathrm{9}}\wedge{y}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$
Answered by behi83417@gmail.com last updated on 05/Feb/20
![(1/x)=t^2 ,(1/y)=s^2 ⇒ { ((t^2 +s^2 =34)),((t+s=23−ts)) :}⇒_(ts=q) ^(t+s=p) { (((t+s)^2 −2ts=34)),((t+s+ts=23)) :} ⇒ { ((p^2 −2q=34)),((p+q=23)) :}⇒p^2 −2(23−p)−34=0 ⇒p^2 +2p+1=34+46+1⇒(p+1)^2 =81 ⇒p+1=±9⇒ { ((p=8⇒q=15)),((p=−10⇒q=33)) :} ⇒ { (([p=8,q=15]⇒[(1/(√x))=3,5,(1/(√y))=5,3])),((⇒(x,y)=((1/9),(1/(25))),((1/(25)),(1/9)))) :} ⇒ { (([p=−10,q=33]⇒z^2 +10z+33=0)),((z=−5±(√(25−4×33))=−5±(√(107))i)) :} ⇒(1/(√x))=−5±(√(107))i⇒ { ((x=(1/((−5−(√(107))i)^2 )))),((y=(1/((−5+(√(107))i)^2 )))) :}](Q80724.png)
$$\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{t}^{\mathrm{2}} ,\frac{\mathrm{1}}{\mathrm{y}}=\mathrm{s}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{t}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} =\mathrm{34}}\\{\mathrm{t}+\mathrm{s}=\mathrm{23}−\mathrm{ts}}\end{cases}\underset{\mathrm{ts}=\mathrm{q}} {\overset{\mathrm{t}+\mathrm{s}=\mathrm{p}} {\Rightarrow}}\:\:\begin{cases}{\left(\mathrm{t}+\mathrm{s}\right)^{\mathrm{2}} −\mathrm{2ts}=\mathrm{34}}\\{\mathrm{t}+\mathrm{s}+\mathrm{ts}=\mathrm{23}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{p}^{\mathrm{2}} −\mathrm{2q}=\mathrm{34}}\\{\mathrm{p}+\mathrm{q}=\mathrm{23}}\end{cases}\Rightarrow\mathrm{p}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{23}−\mathrm{p}\right)−\mathrm{34}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{2}} +\mathrm{2p}+\mathrm{1}=\mathrm{34}+\mathrm{46}+\mathrm{1}\Rightarrow\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{81} \\ $$$$\Rightarrow\mathrm{p}+\mathrm{1}=\pm\mathrm{9}\Rightarrow\begin{cases}{\mathrm{p}=\mathrm{8}\Rightarrow\mathrm{q}=\mathrm{15}}\\{\mathrm{p}=−\mathrm{10}\Rightarrow\mathrm{q}=\mathrm{33}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\left[\mathrm{p}=\mathrm{8},\mathrm{q}=\mathrm{15}\right]\Rightarrow\left[\frac{\mathrm{1}}{\sqrt{\mathrm{x}}}=\mathrm{3},\mathrm{5},\frac{\mathrm{1}}{\sqrt{\mathrm{y}}}=\mathrm{5},\mathrm{3}\right]}\\{\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{\mathrm{1}}{\mathrm{9}},\frac{\mathrm{1}}{\mathrm{25}}\right),\left(\frac{\mathrm{1}}{\mathrm{25}},\frac{\mathrm{1}}{\mathrm{9}}\right)}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\left[\mathrm{p}=−\mathrm{10},\mathrm{q}=\mathrm{33}\right]\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{10z}+\mathrm{33}=\mathrm{0}}\\{\mathrm{z}=−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{4}×\mathrm{33}}=−\mathrm{5}\pm\sqrt{\mathrm{107}}\mathrm{i}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\sqrt{\mathrm{x}}}=−\mathrm{5}\pm\sqrt{\mathrm{107}}\mathrm{i}\Rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{1}}{\left(−\mathrm{5}−\sqrt{\mathrm{107}}\mathrm{i}\right)^{\mathrm{2}} }}\\{\mathrm{y}=\frac{\mathrm{1}}{\left(−\mathrm{5}+\sqrt{\mathrm{107}}\mathrm{i}\right)^{\mathrm{2}} }}\end{cases} \\ $$