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Question Number 8955 by Sopheak last updated on 07/Nov/16

Solve the equation below (√(x−2))=((5x^2 −10x+1)/(x^2 +6x−11))

$${Solve}\:{the}\:{equation}\:{below}\: \\ $$$$\sqrt{{x}−\mathrm{2}}=\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{11}} \\ $$

Commented by prakash jain last updated on 08/Nov/16

x−2=u (√u)=((5(u+2)^2 −10(u+2)+1)/((u+2)^2 +6(u+2)−11)) (√u)=((5u^2 +20u+20−10u−20+1)/(u^2 +4u+4+6u+12−11)) (√u)=((5u^2 +10u+1)/(u^2 +10u+5)) u^2 (√u)−5u^2 +10u(√u)−10u+5(√u)−1=0 (√u)=y y^5 −5y^4 +10y^3 −10y^2 +5y−1=0 (A) y^5 −y^4 −4y^4 +4y^3 +6y^3 −6y^2 −4y^2 +4y+y−1=0 (y−1)(y^4 −4y^3 +6y^2 −4y+1)=0 y=1⇒u=1⇒x=3 sol 1 y^4 −4y^3 +6y^2 −4y+1=0 y^4 −4y^3 +4y^2 +2y^2 −4y+1=0 y^2 (y−2)^2 +2y(y−2)+1=0 (y(y−2)+1)^2 =0 (y^2 −2y+1)^2 =0 (y−1)^4 =0⇒y=1,x=3 all 5 solutions of A are equal y=1⇒u=1⇒x=3

$${x}−\mathrm{2}={u} \\ $$$$\sqrt{{u}}=\frac{\mathrm{5}\left({u}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{10}\left({u}+\mathrm{2}\right)+\mathrm{1}}{\left({u}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}\left({u}+\mathrm{2}\right)−\mathrm{11}} \\ $$$$\sqrt{{u}}=\frac{\mathrm{5}{u}^{\mathrm{2}} +\mathrm{20}{u}+\mathrm{20}−\mathrm{10}{u}−\mathrm{20}+\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{4}{u}+\mathrm{4}+\mathrm{6}{u}+\mathrm{12}−\mathrm{11}} \\ $$$$\sqrt{{u}}=\frac{\mathrm{5}{u}^{\mathrm{2}} +\mathrm{10}{u}+\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{10}{u}+\mathrm{5}} \\ $$$${u}^{\mathrm{2}} \sqrt{{u}}−\mathrm{5}{u}^{\mathrm{2}} +\mathrm{10}{u}\sqrt{{u}}−\mathrm{10}{u}+\mathrm{5}\sqrt{{u}}−\mathrm{1}=\mathrm{0} \\ $$$$\sqrt{{u}}={y} \\ $$$${y}^{\mathrm{5}} −\mathrm{5}{y}^{\mathrm{4}} +\mathrm{10}{y}^{\mathrm{3}} −\mathrm{10}{y}^{\mathrm{2}} +\mathrm{5}{y}−\mathrm{1}=\mathrm{0}\:\:\left(\mathrm{A}\right) \\ $$$${y}^{\mathrm{5}} −{y}^{\mathrm{4}} −\mathrm{4}{y}^{\mathrm{4}} +\mathrm{4}{y}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{3}} −\mathrm{6}{y}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+{y}−\mathrm{1}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}^{\mathrm{4}} −\mathrm{4}{y}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1}\Rightarrow{u}=\mathrm{1}\Rightarrow{x}=\mathrm{3}\:\:{sol}\:\mathrm{1} \\ $$$${y}^{\mathrm{4}} −\mathrm{4}{y}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{4}{y}^{\mathrm{3}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} \left({y}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{y}\left({y}−\mathrm{2}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\left({y}\left({y}−\mathrm{2}\right)+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)^{\mathrm{4}} =\mathrm{0}\Rightarrow{y}=\mathrm{1},{x}=\mathrm{3} \\ $$$$\mathrm{all}\:\mathrm{5}\:\mathrm{solutions}\:\:\mathrm{of}\:{A}\:\mathrm{are}\:\mathrm{equal} \\ $$$${y}=\mathrm{1}\Rightarrow{u}=\mathrm{1}\Rightarrow{x}=\mathrm{3} \\ $$

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