Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 65664 by mathmax by abdo last updated on 01/Aug/19

solve (((√(1−x))−(√(2x+1)))/((√(1−x))+(√(2x+1)))) =((x+1)/3)

$${solve}\:\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}\:=\frac{{x}+\mathrm{1}}{\mathrm{3}} \\ $$

Answered by MJS last updated on 01/Aug/19

(((√(1−x))−(√(2x+1)))/((√(1−x))+(√(2x+1))))=((((√(1−x))−(√(2x+1)))^2 )/(((√(1−x))+(√(2x+1)))((√(1−x))−(√(2x+1)))))=  =((x+2−2(√(1−x))(√(2x+1)))/(−3x))=−((x+2)/(3x))+((2(√(1−x))(√(2x+1)))/(3x))  −((x+2)/(3x))+((2(√(1−x))(√(2x+1)))/(3x))=((x+1)/3)  2(√(1−x))(√(2x+1))=x^2 +2x+2  squaring  x(x^3 +4x^2 +16x+4)=0  x_1 =0 but it′s false  x^3 +4x^2 +16x+4=0  x=t−(4/3)  t=((((170)/(27))+(2/9)(√(1713))))^(1/3) −((−((170)/(27))+(2/9)(√(1713))))^(1/3)   x=−(4/3)+((((170)/(27))+(2/9)(√(1713))))^(1/3) −((−((170)/(27))+(2/9)(√(1713))))^(1/3)   x≈−.266582

$$\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}=\frac{\left(\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)\left(\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)}= \\ $$$$=\frac{{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{−\mathrm{3}{x}}=−\frac{{x}+\mathrm{2}}{\mathrm{3}{x}}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}} \\ $$$$−\frac{{x}+\mathrm{2}}{\mathrm{3}{x}}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}}=\frac{{x}+\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2} \\ $$$$\mathrm{squaring} \\ $$$${x}\left({x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{false} \\ $$$${x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}={t}−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}} \\ $$$${x}=−\frac{\mathrm{4}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}} \\ $$$${x}\approx−.\mathrm{266582} \\ $$

Commented by mathmax by abdo last updated on 01/Aug/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com