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 211761 by MATHEMATICSAM last updated on 20/Sep/24

$$\mathrm{If}x=\frac{4a^{2}b}{4b^2+1}\mathrm{where}b>\frac{1}{2}\mathrm{then}$$$$\frac{\sqrt{a^2+x}+\sqrt{a^2-x}}{\sqrt{a^2+x}-\sqrt{a^2-x}}=?$$

Answered by Rasheed.Sindhi last updated on 21/Sep/24

$$let\frac{p+q}{p-q}=\frac{\sqrt{a^2+x}+\sqrt{a^2-x}}{\sqrt{a^2+x}-\sqrt{a^2-x}}$$$$\frac{p}{q}=\frac{\sqrt{a^2+x}}{\sqrt{a^2-x}}$$$$\frac{p^2}{q^2}=\frac{a^2+x}{a^2-x}$$$$\frac{p^2+q^2}{p^2-q^2}=\frac{a^2}{x}=\frac{a^2}{\frac{4a^{2}b}{4b^2+1}}$$$$=a^2\times \frac{4b^2+1}{4a^{2}b}$$$$=\frac{4b^2+1}{4b}$$$$\frac{p^2}{q^2}=\frac{4b^2+4b+1}{4b^2-4b+1}=\frac{{(2b+1)}^2}{{(2b-1)}^2}$$$$\frac{p}{q}=\frac{2b+1}{2b-1}\mid \frac{p}{q}=\frac{1+2b}{1-2b}$$$$\frac{p+q}{p-q}=\frac{4b}{2}=2b\mid \frac{p+q}{p-q}=\frac{2}{4b}=\frac{1}{2b}$$

Answered by som(math1967) last updated on 21/Sep/24

$$\frac{\sqrt{a^2+x}+\sqrt{a^2-x}}{\sqrt{a^z+x}-\sqrt{a^2-x}}$$$$=\frac{{(\sqrt{a^2+x}+\sqrt{a^2-x})}^2}{{\left(\sqrt{a^2+x}\right)}^2-{\left(\sqrt{a^2-x}\right)}^2}$$$$=\frac{a^2+x+a^2-x+2\sqrt{a^4-x^2}}{a^2+x-a^2+x}$$$$=\frac{2(a^2+\sqrt{a^4-x^2}}{2x}$$$$=\frac{a^2+\sqrt{a^4-\frac{16a^4{b}^2}{{(4b^2+1)}^2}}}{\frac{4a^{2}b}{4b^2+1}}$$$$=\frac{a^2+a^2\sqrt{\frac{{(4b^2+1)}^2-4.4b^2.1}{{(4b^2+1)}^2}}}{\frac{4a^{2}b}{(4b^2+1)}}$$$$=\frac{a^2(1+\frac{4b^2-1}{4b^2+1})}{\frac{4a^{2}b}{4b^2+1}}$$$$=\frac{a^2\times 8b^2}{4b^2+1}\times \frac{4b^2+1}{4a^{2}b}=2b$$$$$$

Commented by som(math1967) last updated on 21/Sep/24

$$yessir,$$

Commented by MATHEMATICSAM last updated on 21/Sep/24

$$\mathrm{sir}\mathrm{the}\mathrm{answer}\mathrm{should}\mathrm{be}2b$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com