Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 155650 by mathdanisur last updated on 03/Oct/21

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$\sqrt{\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}}+\frac{\mathrm{a}}{\mathrm{x}}=\sqrt{\frac{\mathrm{x}-\mathrm{b}}{\mathrm{x}-\mathrm{a}}}+\frac{\mathrm{b}}{\mathrm{x}}$$$$\mathrm{a};\mathrm{b}\in \mathbb{R}\mathrm{and}\mathrm{a}\ne \mathrm{b}$$

Answered by som(math1967) last updated on 03/Oct/21

$$\sqrt{\frac{x-a}{x-b}}-\sqrt{\frac{x-b}{x-a}}=\frac{b-a}{x}$$$$\frac{x-a-x+b}{\sqrt{x^2-(a+b)x+ab}}=\frac{b-a}{x}$$$$\frac{1}{\sqrt{x^2-(a+b)x+ab}}=\frac{1}{x}[\because a\ne b]$$$$x^2-(a+b)x+ab=x^2\left[squaringbothside\right]$$$$x=\frac{ab}{a+b}ans$$$$$$

Commented by Rasheed.Sindhi last updated on 03/Oct/21

$$Youattackafewquestionsbut$$$$youranswersarealwaysperfect!:)$$

Commented by som(math1967) last updated on 03/Oct/21

$$Thankyousir$$

Answered by ajfour last updated on 03/Oct/21

$$\frac{x-a}{x-b}=t^2\Rightarrow x=\frac{{bt}^2-a}{t^2-1}$$$$t-\frac{1}{t}=\frac{(b-a)(t^2-1)}{{bt}^2-a}$$$$ast\ne 1\Rightarrow {bt}^2-a=(b-a)t$$$$x=b+\frac{b-a}{t^2-1}=b+\frac{b(b-a)}{(b-a)t-(b-a)}$$$$x=b+\frac{b}{t-1}$$$$t=\frac{(b-a)}{2b}\pm \sqrt{\frac{{(b-a)}^2}{4b^2}+\frac{a}{b}}$$$$t=\frac{b-a\pm \mid a+b\mid }{2b}$$$$x=b+\frac{2b^2}{-a-b\pm \mid a+b\mid }$$$$rejecting+vesign$$$$\Rightarrow x=b-\frac{b^2}{a+b}$$$$x=\frac{ab}{a+b}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com