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Question Number 8416 by Chantria last updated on 10/Oct/16
Solve equation 1. x^2 +y^2 =x+y+8 (x;y be positive) 2. x^2 −2(√x)+1=0
$$\boldsymbol{{Solve}}\:\boldsymbol{{equation}}\: \\ $$$$\:\mathrm{1}.\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}+\boldsymbol{{y}}+\mathrm{8}\:\:\:\:\:\:\:\left({x};{y}\:{be}\:{positive}\right) \\ $$$$\:\mathrm{2}.\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\sqrt{\boldsymbol{{x}}}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 10/Oct/16
1. x^2 +y^2 =x+y+8 (x;y be positive) 2. x^2 −2(√x)+1=0 x^2 +1=2(√x) (x^2 +1)^2 =(2(√x))^2 x^4 +2x^2 +1=4x x^4 +2x^2 −4x+1=0
$$\:\mathrm{1}.\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}+\boldsymbol{{y}}+\mathrm{8}\:\:\:\:\:\:\:\left({x};{y}\:{be}\:{positive}\right) \\ $$$$\:\mathrm{2}.\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\sqrt{\boldsymbol{{x}}}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{2}\sqrt{\mathrm{x}} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{x}}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{1}=\mathrm{4x} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$
Commented by prakash jain last updated on 11/Oct/16
x^2 −2(√x)+1=0 x^2 −x^(3/2) +x^(3/2) −x+x−(√x)−(√x)+1=0 x^(3/2) ((√x)−1)+x((√x)−1)+(√x)((√x)−1)−1((√x)−1)=0 ((√x)−1)(x^(3/2) +x+(√x)−1)=0 x=1 is one solution u=x^2 −2(√x)+1 u′=2x−(1/( (√x))) u′=((2x^(3/2) −1)/( (√x))) u′<0 for 0<x<1 u′>0 for x>1 u(0)=1 so x=1 is the only solution. x<0 does not satisfy the original equation. equation (1) 1+y^2 =1+y+8 y^2 −y−8=0 y=((1±(√(1^2 +32)))/2) xy>0, x=1⇒y>0 y=((1+(√(33)))/2), x=1
$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{{x}}+\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}^{\mathrm{3}/\mathrm{2}} +{x}^{\mathrm{3}/\mathrm{2}} −{x}+{x}−\sqrt{{x}}−\sqrt{{x}}+\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}/\mathrm{2}} \left(\sqrt{{x}}−\mathrm{1}\right)+{x}\left(\sqrt{{x}}−\mathrm{1}\right)+\sqrt{{x}}\left(\sqrt{{x}}−\mathrm{1}\right)−\mathrm{1}\left(\sqrt{{x}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\mathrm{1}\right)\left({x}^{\mathrm{3}/\mathrm{2}} +{x}+\sqrt{{x}}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\mathrm{is}\:\mathrm{one}\:\mathrm{solution} \\ $$$${u}={x}^{\mathrm{2}} −\mathrm{2}\sqrt{{x}}+\mathrm{1} \\ $$$${u}'=\mathrm{2}{x}−\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$${u}'=\frac{\mathrm{2}{x}^{\mathrm{3}/\mathrm{2}} −\mathrm{1}}{\:\sqrt{{x}}} \\ $$$${u}'<\mathrm{0}\:\mathrm{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${u}'>\mathrm{0}\:\mathrm{for}\:{x}>\mathrm{1} \\ $$$${u}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{so}\:{x}=\mathrm{1}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}. \\ $$$${x}<\mathrm{0}\:\mathrm{does}\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{original} \\ $$$$\mathrm{equation}. \\ $$$${equation}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{1}+{y}^{\mathrm{2}} =\mathrm{1}+{y}+\mathrm{8} \\ $$$${y}^{\mathrm{2}} −{y}−\mathrm{8}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{32}}}{\mathrm{2}} \\ $$$${xy}>\mathrm{0},\:{x}=\mathrm{1}\Rightarrow{y}>\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}},\:{x}=\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 11/Oct/16
I couldn′t even think of it! x^2 −x^(3/2) +x^(3/2) −x+x−(√x)−(√x)+1=0 Is there a general rule? In which situation this apply and how?
$$\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{even}\:\mathrm{think}\:\mathrm{of}\:\mathrm{it}! \\ $$$${x}^{\mathrm{2}} −{x}^{\mathrm{3}/\mathrm{2}} +{x}^{\mathrm{3}/\mathrm{2}} −{x}+{x}−\sqrt{{x}}−\sqrt{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{general}\:\mathrm{rule}?\:\mathrm{In}\:\mathrm{which} \\ $$$$\mathrm{situation}\:\mathrm{this}\:\mathrm{apply}\:\mathrm{and}\:\mathrm{how}? \\ $$$$ \\ $$
Commented by Chantria last updated on 12/Oct/16
Yes, but you have a problem in exercise 1
$$\mathrm{Yes},\:\mathrm{but}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{problem}\:\mathrm{in}\:\mathrm{exercise}\:\mathrm{1} \\ $$
Commented by Chantria last updated on 12/Oct/16
Commented by Rasheed Soomro last updated on 12/Oct/16
Is it given that m and n are whole numbers? (2,3) or (3,2) doesn′t satisfy the second equation.
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{given}\:\mathrm{that}\:\mathrm{m}\:\:\mathrm{and}\:\:\mathrm{n}\:\mathrm{are}\:\mathrm{whole}\:\mathrm{numbers}? \\ $$$$\left(\mathrm{2},\mathrm{3}\right)\:\mathrm{or}\:\left(\mathrm{3},\mathrm{2}\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{second}\:\mathrm{equation}. \\ $$
Commented by prakash jain last updated on 14/Oct/16
I thought that the question is for a system of equation.
$$\mathrm{I}\:\mathrm{thought}\:\mathrm{that}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{for}\:\mathrm{a}\: \\ $$$$\mathrm{system}\:\mathrm{of}\:\mathrm{equation}. \\ $$

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