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Question Number 105927 by bemath last updated on 01/Aug/20
((x/(x−2)))^2 +((x/(x+2)))^2 = 2
$$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$
Commented by Rasheed.Sindhi last updated on 02/Aug/20
((x/(x−2)))^2 +((x/(x+2)))^2 = 2 ((x/(x−2)))^2 +((x/(x+2)))^2 = ((√2) )^2 Dividing by ((√2))^2 : ((x/( (√2)(x−2))))^2 +((x/( (√2)(x+2))))^2 = 1 (x^2 /(2(x−2)^2 ))+(x^2 /(2(x+2)^2 ))−1=0 (x^2 /(2(x−2)^2 ))+((x^2 −2x^2 −8x−8)/(2(x+2)^2 ))=0 (x^2 /((x−2)^2 ))−((x^2 +8x+8)/((x+2)^2 ))=0 x^2 (x+2)^2 −(x−2)^2 (x^2 +8x+8)=0 x^4 +4x^3 +4x^2 −(x^2 −4x+4)(x^2 +8x+8)=0 24x^2 −32=0 x^2 =((32)/(24))=(4/3) x=±(2/( (√3)))
$$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$$$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$${Dividing}\:{by}\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} : \\ $$$$\left(\frac{{x}}{\:\sqrt{\mathrm{2}}\left({x}−\mathrm{2}\right)}\right)^{\mathrm{2}} +\left(\frac{{x}}{\:\sqrt{\mathrm{2}}\left({x}+\mathrm{2}\right)}\right)^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{2}\left({x}+\mathrm{2}\right)^{\mathrm{2}} }−\mathrm{1}=\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{8}}{\mathrm{2}\left({x}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{8}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left({x}−\mathrm{2}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{8}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\mathrm{24}{x}^{\mathrm{2}} −\mathrm{32}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{32}}{\mathrm{24}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}=\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by bemath last updated on 02/Aug/20
thank you both
$${thank}\:{you}\:{both}\: \\ $$
Answered by bobhans last updated on 01/Aug/20
Commented by Rasheed.Sindhi last updated on 02/Aug/20
Slightly differentfrom above ((x/(x−2))−(x/(x+2)))^2 +2((x/(x−2)))((x/(x+2)))=2 (((x^2 +2x−x^2 +2x)/(x−2)))^2 +2((x^2 /(x^2 −4)))=2 (((4x)/(x^2 −4)))^2 +2((x^2 /(x^2 −4)))=2 16x^2 +2(((x^2 (x^2 −4)^2 )/(x^2 −4)))=2(x^2 −4)^2 16x^2 +2x^4 ^(×) −8x^2 =2x^4 ^(×) −16x^2 +32 24x^2 =32 x^2 =((32)/(24))=(4/3) x=±(2/( (√3)))
$$\boldsymbol{{Slightly}}\:\boldsymbol{{different}}{from}\:{above} \\ $$$$\left(\frac{{x}}{{x}−\mathrm{2}}−\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{x}}{{x}−\mathrm{2}}\right)\left(\frac{{x}}{{x}+\mathrm{2}}\right)=\mathrm{2} \\ $$$$\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{2}{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{4}}\right)=\mathrm{2} \\ $$$$\left(\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{4}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{4}}\right)=\mathrm{2} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{2}\left(\frac{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{4}}\right)=\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\overset{×} {\mathrm{2}{x}^{\mathrm{4}} }−\mathrm{8}{x}^{\mathrm{2}} =\overset{×} {\mathrm{2}{x}^{\mathrm{4}} }−\mathrm{16}{x}^{\mathrm{2}} +\mathrm{32} \\ $$$$\mathrm{24}{x}^{\mathrm{2}} =\mathrm{32} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{32}}{\mathrm{24}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}=\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$
Answered by Rasheed.Sindhi last updated on 02/Aug/20
((x/(x−2)))^2 +((x/(x+2)))^2 = 2 ((1/((x−2)/x)))^2 +((1/((x+2)/x)))^2 =2 ((1/(1−(2/x))))^2 +((1/(1+(2/x))))^2 =2 ((1/(1−y)))^2 +((1/(1+y)))^2 =2 ; (2/x)=y (1/((1−y)^2 ))+(1/((1+y)^2 ))=2 (1+y)^2 +(1−y)^2 =2(1−y^2 )^2 2(1+y^2 )=2(1−y^2 )^2 y^4 −2y^2 +1=1+y^2 y^4 −3y^2 =0 y^2 (y^2 −3)=0 y=0 ∨ y=±(√3) (2/x)=0 (no solution) ∨ (2/x)=±(√3) x=∞_(wrong approach) ^(×) ∨ x=±(2/( (√3)))
$$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$$$\left(\frac{\mathrm{1}}{\frac{{x}−\mathrm{2}}{{x}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\frac{{x}+\mathrm{2}}{{x}}}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}\right)^{\mathrm{2}} =\mathrm{2}\:\: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}−{y}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{1}+{y}}\right)^{\mathrm{2}} =\mathrm{2}\:;\:\frac{\mathrm{2}}{{x}}={y} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$$\left(\mathrm{1}+{y}\right)^{\mathrm{2}} +\left(\mathrm{1}−{y}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=\mathrm{2}\left(\mathrm{1}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{1}+{y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{4}} −\mathrm{3}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}^{\mathrm{2}} \left({y}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{0} \\ $$$${y}=\mathrm{0}\:\vee\:{y}=\pm\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{2}}{{x}}=\mathrm{0}\:\left({no}\:{solution}\right)\:\vee\:\frac{\mathrm{2}}{{x}}=\pm\sqrt{\mathrm{3}} \\ $$$$\underset{{wrong}\:{approach}} {\overset{×} {{x}=\infty}}\:\vee\:{x}=\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by bobhans last updated on 01/Aug/20
waw....if x= ∞ then ((∞/(∞−2)))^2 +((∞/(∞+2)))^2 =2 it is true sir?
$$\mathrm{waw}….\mathrm{if}\:\mathrm{x}=\:\infty\:\mathrm{then}\:\left(\frac{\infty}{\infty−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\infty}{\infty+\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{sir}?\: \\ $$
Commented by Rasheed.Sindhi last updated on 01/Aug/20
Sir I didn′t try to check the equation back. Perhaps this is what you have said “nothing solution”.I think ′nothing solution′ is better. So , (2/x)=0⇒2=0(meaningless) so no solution. BTW (2/x)=0 and also (2/∞)=0 I concluded that x=∞
$$\mathcal{S}{ir}\:{I}\:{didn}'{t}\:{try}\:{to}\:{check}\:{the} \\ $$$${equation}\:{back}.\:{Perhaps}\:{this}\:{is} \\ $$$${what}\:{you}\:{have}\:{said}\:“{nothing} \\ $$$${solution}''.{I}\:{think}\:'{nothing} \\ $$$${solution}'\:{is}\:\:{better}. \\ $$$${So}\:,\:\:\:\:\:\frac{\mathrm{2}}{{x}}=\mathrm{0}\Rightarrow\mathrm{2}=\mathrm{0}\left({meaningless}\right) \\ $$$${so}\:{no}\:{solution}. \\ $$$${BTW}\: \\ $$$$\:\:\:\:\frac{\mathrm{2}}{{x}}=\mathrm{0}\:\:\:{and}\:{also}\:\:\:\frac{\mathrm{2}}{\infty}=\mathrm{0} \\ $$$${I}\:{concluded}\:{that}\:{x}=\infty \\ $$
Commented by bobhans last updated on 02/Aug/20
ok sir. thank you
$$\mathrm{ok}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Rasheed.Sindhi last updated on 02/Aug/20
((x/(x−2)))^2 +((x/(x+2)))^2 = 2 Dividing by ((x/(x−2)))^2 1+(((x/(x+2))/(x/(x−2))))^2 = (2/(((x/(x−2)))^2 )) 1+(((x−2)/(x+2)))^2 =((2(x−2)^2 )/x^2 ) (((x+2)^2 +(x−2)^2 )/((x+2)^2 ))=((2(x−2)^2 )/x^2 ) ((2(x^2 +4))/((x+2)^2 ))=((2(x−2)^2 )/x^2 ) x^2 (x^2 +4)=(x^2 −4)^2 x^4 +4x^2 =x^4 −8x^2 +16 12x^2 =16 x=±(2/( (√3)))
$$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$$${Dividing}\:{by}\:\:\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\left(\frac{\frac{{x}}{{x}+\mathrm{2}}}{\frac{{x}}{{x}−\mathrm{2}}}\right)^{\mathrm{2}} \:\:=\:\:\frac{\mathrm{2}}{\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{1}+\left(\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)=\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} ={x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{16} \\ $$$$\mathrm{12}{x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:{x}=\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Aug/20
Similarly the question can be solved also by dividing ((x/(x+2)))^2 to both sides.
$$\mathcal{S}{imilarly}\:{the}\:{question}\:{can}\:{be}\:{solved} \\ $$$${also}\:{by}\:{dividing}\:\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} {to}\:{both} \\ $$$${sides}. \\ $$

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