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Question Number 215559 by hardmath last updated on 10/Jan/25

(√y) + (√x) = 5 (√x) ∙ (√y) = 8 (((√x) y − x (√y))/(y − x)) = ?

$$\sqrt{\mathrm{y}}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\sqrt{\mathrm{x}}\:\centerdot\:\sqrt{\mathrm{y}}\:=\:\mathrm{8} \\ $$$$\frac{\sqrt{\mathrm{x}}\:\mathrm{y}\:−\:\mathrm{x}\:\sqrt{\mathrm{y}}}{\mathrm{y}\:−\:\mathrm{x}}\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 10/Jan/25

(√y) +(√x) =5 ; (√x) (√y) =8 (((√x) y − x (√y))/(y − x)) =(((√(xy)) ((√y) −(√x) ))/(((√y) )^2 −((√x) )^2 )) = (((√(xy)) ((√y) −(√x) ))/(((√y) −(√x) )((√y) +(√x) )))=(8/5)

$$\sqrt{\mathrm{y}}\:+\sqrt{\mathrm{x}}\:=\mathrm{5}\:;\:\sqrt{\mathrm{x}}\:\sqrt{\mathrm{y}}\:=\mathrm{8} \\ $$$$\frac{\sqrt{\mathrm{x}}\:\mathrm{y}\:−\:\mathrm{x}\:\sqrt{\mathrm{y}}}{\mathrm{y}\:−\:\mathrm{x}} \\ $$$$=\frac{\sqrt{\mathrm{xy}}\:\left(\sqrt{\mathrm{y}}\:−\sqrt{\mathrm{x}}\:\right)}{\left(\sqrt{\mathrm{y}}\:\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{x}}\:\right)^{\mathrm{2}} } \\ $$$$\:=\:\frac{\sqrt{\mathrm{xy}}\:\cancel{\left(\sqrt{\mathrm{y}}\:−\sqrt{\mathrm{x}}\:\right)}}{\cancel{\left(\sqrt{\mathrm{y}}\:−\sqrt{\mathrm{x}}\:\right)}\left(\sqrt{\mathrm{y}}\:+\sqrt{\mathrm{x}}\:\right)}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$

Commented by Frix last updated on 11/Jan/25

Yes!

$$\mathrm{Yes}! \\ $$

Answered by Frix last updated on 11/Jan/25

To find x, y with α, β ∈R A (√x)+(√y)=α B (√x)(√y)=β Let (√x)=a+bi∧(√y)=a−bi; b≥0 A 2a=α B a^2 +b^2 =β ========= A a=(α/2) B b=(√(β−(α^2 /4))) ========= ⇒ (√x)=(α/2)+((√(4β−α^2 ))/2)i (√y)=(α/2)−((√(4β−α^2 ))/2)i ⇒ x=((α^2 −2β)/2)+((α(√(4β−α^2 )))/2)i y=((α^2 −2β)/2)−((α(√(4β−α^2 )))/2)i x, y ∈R ⇔ β≤(α^2 /4) x=βe^(i cos^(−1) ((α^2 −2β)/(2β)) ) y=βe^(−i cos^(−1) ((α^2 −2β)/(2β)) )

$$\mathrm{To}\:\mathrm{find}\:{x},\:{y}\:\mathrm{with}\:\alpha,\:\beta\:\in\mathbb{R} \\ $$$${A}\:\:\:\:\:\sqrt{{x}}+\sqrt{{y}}=\alpha \\ $$$${B}\:\:\:\:\:\sqrt{{x}}\sqrt{{y}}=\beta \\ $$$$\mathrm{Let}\:\sqrt{{x}}={a}+{b}\mathrm{i}\wedge\sqrt{{y}}={a}−{b}\mathrm{i};\:{b}\geqslant\mathrm{0} \\ $$$${A}\:\:\:\:\:\mathrm{2}{a}=\alpha \\ $$$${B}\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\beta \\ $$$$========= \\ $$$${A}\:\:\:\:\:{a}=\frac{\alpha}{\mathrm{2}} \\ $$$${B}\:\:\:\:\:{b}=\sqrt{\beta−\frac{\alpha^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$========= \\ $$$$\Rightarrow \\ $$$$\sqrt{{x}}=\frac{\alpha}{\mathrm{2}}+\frac{\sqrt{\mathrm{4}\beta−\alpha^{\mathrm{2}} }}{\mathrm{2}}\mathrm{i} \\ $$$$\sqrt{{y}}=\frac{\alpha}{\mathrm{2}}−\frac{\sqrt{\mathrm{4}\beta−\alpha^{\mathrm{2}} }}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\alpha^{\mathrm{2}} −\mathrm{2}\beta}{\mathrm{2}}+\frac{\alpha\sqrt{\mathrm{4}\beta−\alpha^{\mathrm{2}} }}{\mathrm{2}}\mathrm{i} \\ $$$${y}=\frac{\alpha^{\mathrm{2}} −\mathrm{2}\beta}{\mathrm{2}}−\frac{\alpha\sqrt{\mathrm{4}\beta−\alpha^{\mathrm{2}} }}{\mathrm{2}}\mathrm{i} \\ $$$${x},\:{y}\:\in\mathbb{R}\:\Leftrightarrow\:\beta\leqslant\frac{\alpha^{\mathrm{2}} }{\mathrm{4}} \\ $$$${x}=\beta\mathrm{e}^{\mathrm{i}\:\mathrm{cos}^{−\mathrm{1}} \:\frac{\alpha^{\mathrm{2}} −\mathrm{2}\beta}{\mathrm{2}\beta}\:} \\ $$$${y}=\beta\mathrm{e}^{−\mathrm{i}\:\mathrm{cos}^{−\mathrm{1}} \:\frac{\alpha^{\mathrm{2}} −\mathrm{2}\beta}{\mathrm{2}\beta}\:} \\ $$

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