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Question Number 154036 by liberty last updated on 13/Sep/21

49(((x+5)/(x−2)))^2 +36(((x+5)/(x−1)))^2 = 85

$$\:\mathrm{49}\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{36}\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\:\mathrm{85} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Sep/21

49(((x+5)/(x−2)))^2 +36(((x+5)/(x−1)))^2 = 85 85=36+49=6^2 +7^2 85=4+81=2^2 +9^2 Four cases: (((7(x+5))/(x−2)))^2 +(((6(x+5))/(x−1)))^2 = 6^2 +7^2 ...(1) (((7(x+5))/(x−2)))^2 +(((6(x+5))/(x−1)))^2 = 7^2 +6^2 ...(2) (((7(x+5))/(x−2)))^2 +(((6(x+5))/(x−1)))^2 = 2^2 +9^2 ...(3) (((7(x+5))/(x−2)))^2 +(((6(x+5))/(x−1)))^2 = 9^2 +2^2 ...(4) (1) ((x+5)/(x−2))=±(6/7) ∧ ((x+5)/(x−1))=±(7/6)^★ 7x+35=±6x∓12 7x∓6x=∓12−35 (x,13x)=(−47,−23) (x,x)=(−47,((−23)/(13))) ^★ 6x+30=±7x∓7 6x∓7x=∓7−30 (−x,13x=−37,−23 (x=37,x=((−23)/(13))) x=−((23)/(13)) .....

$$ \\ $$$$\:\mathrm{49}\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{36}\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\:\mathrm{85} \\ $$$$\:\:\:\mathrm{85}=\mathrm{36}+\mathrm{49}=\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{85}=\mathrm{4}+\mathrm{81}=\mathrm{2}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} \\ $$$${Four}\:{cases}: \\ $$$$\left(\frac{\mathrm{7}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\:\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} ...\left(\mathrm{1}\right) \\ $$$$\left(\frac{\mathrm{7}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\:\mathrm{7}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} ...\left(\mathrm{2}\right) \\ $$$$\left(\frac{\mathrm{7}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\:\mathrm{2}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} ...\left(\mathrm{3}\right) \\ $$$$\left(\frac{\mathrm{7}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\:\mathrm{9}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} ...\left(\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1}\right)\:\:\:\:\:\frac{{x}+\mathrm{5}}{{x}−\mathrm{2}}=\pm\frac{\mathrm{6}}{\mathrm{7}}\:\wedge\:\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}=\pm\frac{\mathrm{7}}{\mathrm{6}}\:^{\bigstar} \\ $$$$\:\:\:\:\mathrm{7}{x}+\mathrm{35}=\pm\mathrm{6}{x}\mp\mathrm{12} \\ $$$$\:\:\:\mathrm{7}{x}\mp\mathrm{6}{x}=\mp\mathrm{12}−\mathrm{35} \\ $$$$\:\:\:\left({x},\mathrm{13}{x}\right)=\left(−\mathrm{47},−\mathrm{23}\right) \\ $$$$\left({x},{x}\right)=\left(−\mathrm{47},\frac{−\mathrm{23}}{\mathrm{13}}\right) \\ $$$$\:^{\bigstar} \mathrm{6}{x}+\mathrm{30}=\pm\mathrm{7}{x}\mp\mathrm{7} \\ $$$$\mathrm{6}{x}\mp\mathrm{7}{x}=\mp\mathrm{7}−\mathrm{30} \\ $$$$\left(−{x},\mathrm{13}{x}=−\mathrm{37},−\mathrm{23}\right. \\ $$$$\left({x}=\mathrm{37},{x}=\frac{−\mathrm{23}}{\mathrm{13}}\right) \\ $$$${x}=−\frac{\mathrm{23}}{\mathrm{13}} \\ $$$$..... \\ $$

Commented by iloveisrael last updated on 14/Sep/21

waw....amazing

$${waw}....{amazing} \\ $$

Commented by Rasheed.Sindhi last updated on 14/Sep/21

ThαnX sir!

$$\mathcal{T}{h}\alpha{n}\mathcal{X}\:\:{sir}! \\ $$

Answered by Rasheed.Sindhi last updated on 13/Sep/21

(((7(x+5))/(x−2)))^2 _(u^2 ) +(((6(x+5))/(x−1)))^2 _(v^2 ) = 85 u^2 +v^2 =85 ((7(x+5))/(x−2))=u 7x+35=ux−2u⇒ux−7x=2u+35 ⇒x=((2u+35)/(u−7)) v=((6(x+5))/(x−1))⇒6x+30=vx−v x(v−6)=v+30 (((2u+35)/(u−7)))(v−6)=v+30 v(((2u+35)/(u−7)))−6(((2u+35)/(u−7)))=v+30 v(((2u+35)/(u−7)))−v=6(((2u+35)/(u−7)))+30 v(((2u+35)/(u−7))−1)=((12u+210+30u−210)/(u−7)) v(((2u+35−u+7)/(u−7)))=((12u+210+30u−210)/(u−7)) v(((u+42)/(u−7)))=((42u)/(u−7)) v=((42u)/(u+42)) u^2 +v^2 =85⇒u^2 +(((42u)/(u+42)))^2 =85 u^2 (u+42)^2 +42^2 u^2 −85(u+42)^2 =0 u^2 (u^2 +84u+42^2 )+42^2 u^2 −85(u^2 +84u+42^2 )=0 u^4 +84u^3 +42^2 u^2 +42^2 u^2 −85u^2 −7140u−85(42^2 )=0 u^4 +84u^3 +(2.42^2 −85)u^2 −7140u−85(42^2 )=0 ... ...... ....

$$\:\underset{\mathrm{u}^{\mathrm{2}} } {\underbrace{\left(\frac{\mathrm{7}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{2}}\right)^{\mathrm{2}} }}+\underset{\mathrm{v}^{\mathrm{2}} } {\underbrace{\left(\frac{\mathrm{6}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{1}}\right)^{\mathrm{2}} }}=\:\mathrm{85} \\ $$$$\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{85} \\ $$$$\frac{\mathrm{7}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{2}}=\mathrm{u} \\ $$$$\mathrm{7}{x}+\mathrm{35}={ux}−\mathrm{2}{u}\Rightarrow{ux}−\mathrm{7}{x}=\mathrm{2}{u}+\mathrm{35} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}} \\ $$$$\mathrm{v}=\frac{\mathrm{6}\left({x}+\mathrm{5}\right)}{{x}−\mathrm{1}}\Rightarrow\mathrm{6}{x}+\mathrm{30}=\mathrm{v}{x}−\mathrm{v} \\ $$$${x}\left(\mathrm{v}−\mathrm{6}\right)=\mathrm{v}+\mathrm{30} \\ $$$$\left(\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}}\right)\left(\mathrm{v}−\mathrm{6}\right)=\mathrm{v}+\mathrm{30} \\ $$$$\mathrm{v}\left(\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}}\right)−\mathrm{6}\left(\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}}\right)=\mathrm{v}+\mathrm{30} \\ $$$$\mathrm{v}\left(\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}}\right)−\mathrm{v}=\mathrm{6}\left(\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}}\right)+\mathrm{30} \\ $$$$\mathrm{v}\left(\frac{\mathrm{2}{u}+\mathrm{35}}{{u}−\mathrm{7}}−\mathrm{1}\right)=\frac{\mathrm{12u}+\cancel{\mathrm{210}}+\mathrm{30u}−\cancel{\mathrm{210}}}{\mathrm{u}−\mathrm{7}} \\ $$$$\mathrm{v}\left(\frac{\mathrm{2}{u}+\mathrm{35}−\mathrm{u}+\mathrm{7}}{{u}−\mathrm{7}}\right)=\frac{\mathrm{12u}+\mathrm{210}+\mathrm{30u}−\mathrm{210}}{\mathrm{u}−\mathrm{7}} \\ $$$$\mathrm{v}\left(\frac{{u}+\mathrm{42}}{{u}−\mathrm{7}}\right)=\frac{\mathrm{42u}}{\mathrm{u}−\mathrm{7}} \\ $$$$\mathrm{v}=\frac{\mathrm{42u}}{\mathrm{u}+\mathrm{42}} \\ $$$$\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{85}\Rightarrow\mathrm{u}^{\mathrm{2}} +\left(\frac{\mathrm{42u}}{\mathrm{u}+\mathrm{42}}\right)^{\mathrm{2}} =\mathrm{85} \\ $$$$\mathrm{u}^{\mathrm{2}} \left(\mathrm{u}+\mathrm{42}\right)^{\mathrm{2}} +\mathrm{42}^{\mathrm{2}} \mathrm{u}^{\mathrm{2}} −\mathrm{85}\left(\mathrm{u}+\mathrm{42}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{2}} \left(\mathrm{u}^{\mathrm{2}} +\mathrm{84u}+\mathrm{42}^{\mathrm{2}} \right)+\mathrm{42}^{\mathrm{2}} \mathrm{u}^{\mathrm{2}} −\mathrm{85}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{84u}+\mathrm{42}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{4}} +\mathrm{84u}^{\mathrm{3}} +\mathrm{42}^{\mathrm{2}} \mathrm{u}^{\mathrm{2}} +\mathrm{42}^{\mathrm{2}} \mathrm{u}^{\mathrm{2}} −\mathrm{85u}^{\mathrm{2}} −\mathrm{7140u}−\mathrm{85}\left(\mathrm{42}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{4}} +\mathrm{84u}^{\mathrm{3}} +\left(\mathrm{2}.\mathrm{42}^{\mathrm{2}} −\mathrm{85}\right)\mathrm{u}^{\mathrm{2}} −\mathrm{7140u}−\mathrm{85}\left(\mathrm{42}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$... \\ $$$$...... \\ $$$$.... \\ $$

Answered by MJS_new last updated on 13/Sep/21

49(((x+5)/(x−2)))^2 +36(((x+5)/(x−1)))^2 −85=0 ((1118x^3 −1207x^2 −3100x+4485)/((x−2)^2 (x−1)^2 ))=0 (x+((23)/(13)))(x^2 −((245)/(86))+((195)/(86)))=0 x=−((23)/(13))∨x=((245±(√(7055))i)/(172))

$$\mathrm{49}\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{36}\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{85}=\mathrm{0} \\ $$$$\frac{\mathrm{1118}{x}^{\mathrm{3}} −\mathrm{1207}{x}^{\mathrm{2}} −\mathrm{3100}{x}+\mathrm{4485}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{23}}{\mathrm{13}}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{245}}{\mathrm{86}}+\frac{\mathrm{195}}{\mathrm{86}}\right)=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{23}}{\mathrm{13}}\vee{x}=\frac{\mathrm{245}\pm\sqrt{\mathrm{7055}}\mathrm{i}}{\mathrm{172}} \\ $$

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