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Question Number 194915 by York12 last updated on 19/Jul/23

(3/(x−3))+(5/(x−5))+(7/(x−17))+((19)/(x−19))=x^2 −11x−4

$$ \\ $$$$\frac{\mathrm{3}}{{x}−\mathrm{3}}+\frac{\mathrm{5}}{{x}−\mathrm{5}}+\frac{\mathrm{7}}{{x}−\mathrm{17}}+\frac{\mathrm{19}}{{x}−\mathrm{19}}={x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{4} \\ $$

Commented by BaliramKumar last updated on 19/Jul/23

typo 7→17

$$\mathrm{typo}\:\:\:\mathrm{7}\rightarrow\mathrm{17} \\ $$

Commented by York12 last updated on 19/Jul/23

no sir , I am sure

$${no}\:{sir}\:,\:{I}\:{am}\:{sure} \\ $$

Commented by BaliramKumar last updated on 19/Jul/23

i guess just

$$\mathrm{i}\:\mathrm{guess}\:\mathrm{just} \\ $$

Commented by justenspi last updated on 19/Jul/23

8sybe it is a typo in my book

$$\mathrm{8}{sybe}\:{it}\:{is}\:{a}\:{typo}\:{in}\:{my}\:{book} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Jul/23

(3/(x−3))+(5/(x−5))+((17)/(x−17))+((19)/(x−19))=x^2 −11x−4 (3/(x−3))+1+(5/(x−5))+1+((17)/(x−17))+1+((19)/(x−19))+1=x^2 −11x (x/(x−3))+(x/(x−5))+(x/(x−17))+(x/(x−19))−x^2 +11x=0 x((1/(x−3))+(1/(x−5))+(1/(x−17))+(1/(x−19))−x+11)=0 { ((x=0 ✓)),(((1/(x−3))+(1/(x−19))+(1/(x−5))+(1/(x−17))−x+11=0)) :} ((2x−22)/(x^2 −22x+57))+((2x−22)/(x^2 −22x+85))−(x−11)=0 ((2(x−11))/(x^2 −22x+57))+((2(x−11))/(x^2 −22x+85))−(x−11)=0 (x−11)((2/(x^2 −22x+57))+(2/(x^2 −22x+85))−1)=0 { ((x−11=0⇒x=11✓)),(((2/(x^2 −22x+57))+(2/(x^2 −22x+85))−1=0)) :} (2/(x^2 −22x+57))+(2/(x^2 −22x+85))=1 Let x^2 −22x=y (2/(y+57))+(2/(y+85))=1 2y+170+2y+114=y^2 +142y+4845 y^2 +138y+4561=0 y=−69±10(√2) x^2 −22x=−69±10(√2) ...... ...........

$$\frac{\mathrm{3}}{{x}−\mathrm{3}}+\frac{\mathrm{5}}{{x}−\mathrm{5}}+\frac{\mathrm{17}}{{x}−\mathrm{17}}+\frac{\mathrm{19}}{{x}−\mathrm{19}}={x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{4} \\ $$$$\frac{\mathrm{3}}{{x}−\mathrm{3}}+\mathrm{1}+\frac{\mathrm{5}}{{x}−\mathrm{5}}+\mathrm{1}+\frac{\mathrm{17}}{{x}−\mathrm{17}}+\mathrm{1}+\frac{\mathrm{19}}{{x}−\mathrm{19}}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{11}{x} \\ $$$$\frac{{x}}{{x}−\mathrm{3}}+\frac{{x}}{{x}−\mathrm{5}}+\frac{{x}}{{x}−\mathrm{17}}+\frac{{x}}{{x}−\mathrm{19}}−{x}^{\mathrm{2}} +\mathrm{11}{x}=\mathrm{0} \\ $$$${x}\left(\frac{\mathrm{1}}{{x}−\mathrm{3}}+\frac{\mathrm{1}}{{x}−\mathrm{5}}+\frac{\mathrm{1}}{{x}−\mathrm{17}}+\frac{\mathrm{1}}{{x}−\mathrm{19}}−{x}+\mathrm{11}\right)=\mathrm{0} \\ $$$$\begin{cases}{{x}=\mathrm{0}\:\checkmark}\\{\frac{\mathrm{1}}{{x}−\mathrm{3}}+\frac{\mathrm{1}}{{x}−\mathrm{19}}+\frac{\mathrm{1}}{{x}−\mathrm{5}}+\frac{\mathrm{1}}{{x}−\mathrm{17}}−{x}+\mathrm{11}=\mathrm{0}}\end{cases} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{22}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}{x}−\mathrm{22}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\left({x}−\mathrm{11}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{2}\left({x}−\mathrm{11}\right)}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}\left({x}−\mathrm{11}\right)}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\left({x}−\mathrm{11}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{11}\right)\left(\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{cases}{{x}−\mathrm{11}=\mathrm{0}\Rightarrow{x}=\mathrm{11}\checkmark}\\{\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$$\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}=\mathrm{1} \\ $$$${Let}\:{x}^{\mathrm{2}} −\mathrm{22}{x}={y} \\ $$$$\frac{\mathrm{2}}{{y}+\mathrm{57}}+\frac{\mathrm{2}}{{y}+\mathrm{85}}=\mathrm{1} \\ $$$$\mathrm{2}{y}+\mathrm{170}+\mathrm{2}{y}+\mathrm{114}={y}^{\mathrm{2}} +\mathrm{142}{y}+\mathrm{4845} \\ $$$${y}^{\mathrm{2}} +\mathrm{138}{y}+\mathrm{4561}=\mathrm{0} \\ $$$$\:\:\:{y}=−\mathrm{69}\pm\mathrm{10}\sqrt{\mathrm{2}}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{22}{x}=−\mathrm{69}\pm\mathrm{10}\sqrt{\mathrm{2}}\: \\ $$$$\:\:\:...... \\ $$$$........... \\ $$

Commented by York12 last updated on 19/Jul/23

tbanks

$${tbanks} \\ $$

Commented by BaliramKumar last updated on 19/Jul/23

Nice approach

$$\mathrm{Nice}\:\mathrm{approach} \\ $$

Commented by Rasheed.Sindhi last updated on 20/Jul/23

🙏

🙏

Answered by Rasheed.Sindhi last updated on 20/Jul/23

(3/(x−3))+((19)/(x−19))+(5/(x−5))+((17)/(x−17))=x^2 −11x−4 ((3x−57+19x−57)/(x^2 −22x+57))+((5x−85+17x−85)/(x^2 −22x+85))=x^2 −11x−4 ((22x−114)/(x^2 −22x+57))+2+((22x−170)/(x^2 −22x+85))+2=x^2 −11x ((2x^2 −22x)/(x^2 −22x+57))+((2x^2 −22x)/(x^2 −22x+85))−(x−11)x=0 x(x−11)((2/(x^2 −22x+57))+(2/(x^2 −22x+85))−1)=0 x=0,x=11, (2/(x^2 −22x+57))+(2/(x^2 −22x+85))−1=0 Let x^2 −22x=y (2/(y+57))+(2/(y+85))=1 2y+170+2y+114=y^2 +142y+4845 y^2 +138y+4561=0 y=−69±10(√2) x^2 −22x=−69±10(√2)

$$\frac{\mathrm{3}}{{x}−\mathrm{3}}+\frac{\mathrm{19}}{{x}−\mathrm{19}}+\frac{\mathrm{5}}{{x}−\mathrm{5}}+\frac{\mathrm{17}}{{x}−\mathrm{17}}={x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{4} \\ $$$$\frac{\mathrm{3}{x}−\mathrm{57}+\mathrm{19}{x}−\mathrm{57}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{5}{x}−\mathrm{85}+\mathrm{17}{x}−\mathrm{85}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}={x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{4} \\ $$$$\frac{\mathrm{22}{x}−\mathrm{114}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\mathrm{2}+\frac{\mathrm{22}{x}−\mathrm{170}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}+\mathrm{2}={x}^{\mathrm{2}} −\mathrm{11}{x} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{22}{x}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{22}{x}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\left({x}−\mathrm{11}\right){x}=\mathrm{0} \\ $$$${x}\left({x}−\mathrm{11}\right)\left(\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0},{x}=\mathrm{11},\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{57}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{85}}−\mathrm{1}=\mathrm{0} \\ $$$${Let}\:{x}^{\mathrm{2}} −\mathrm{22}{x}={y} \\ $$$$\frac{\mathrm{2}}{{y}+\mathrm{57}}+\frac{\mathrm{2}}{{y}+\mathrm{85}}=\mathrm{1} \\ $$$$\mathrm{2}{y}+\mathrm{170}+\mathrm{2}{y}+\mathrm{114}={y}^{\mathrm{2}} +\mathrm{142}{y}+\mathrm{4845} \\ $$$${y}^{\mathrm{2}} +\mathrm{138}{y}+\mathrm{4561}=\mathrm{0} \\ $$$${y}=−\mathrm{69}\pm\mathrm{10}\sqrt{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{22}{x}=−\mathrm{69}\pm\mathrm{10}\sqrt{\mathrm{2}} \\ $$

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