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If-x-4a-2-b-4b-2-1-where-b-gt-1-2-then-a-2-x-a-2-x-a-2-x-a-2-x-

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Question Number 211761 by MATHEMATICSAM last updated on 20/Sep/24
If x = ((4a^2 b)/(4b^2 + 1)) where b > (1/2) then (((√(a^2 + x)) + (√(a^2 − x)))/( (√(a^2 + x)) − (√(a^2 − x)))) = ?
$$\mathrm{If}\:{x}\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} {b}}{\mathrm{4}{b}^{\mathrm{2}} \:+\:\mathrm{1}}\:\mathrm{where}\:{b}\:>\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{then} \\ $$$$\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{x}}\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{x}}}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{x}}\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{x}}}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 21/Sep/24
let ((p+q)/(p−q))=(((√(a^2 + x)) + (√(a^2 − x)))/( (√(a^2 + x)) − (√(a^2 − x)))) (p/q)=((√(a^2 +x))/( (√(a^2 −x)))) (p^2 /q^2 )=((a^2 +x)/( a^2 −x)) ((p^2 +q^2 )/(p^2 −q^2 ))=(a^2 /( x))=(a^2 /((4a^2 b)/(4b^2 + 1))) =a^2 ×((4b^2 +1)/(4a^2 b)) =((4b^2 +1)/(4b)) (p^2 /q^2 )=((4b^2 +4b+1)/(4b^2 −4b+1))=(((2b+1)^2 )/((2b−1)^2 )) (p/q)=((2b+1)/(2b−1)) ∣ (p/q)=((1+2b)/(1−2b)) ((p+q)/(p−q))=((4b)/2) =2b∣ ((p+q)/(p−q))=(2/(4b))=(1/(2b))
$${let}\:\:\frac{{p}+{q}}{{p}−{q}}=\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{x}}\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{x}}}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{x}}\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{x}}} \\ $$$$\frac{{p}}{{q}}=\frac{\sqrt{{a}^{\mathrm{2}} +{x}}}{\:\sqrt{{a}^{\mathrm{2}} −{x}}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} +{x}}{\:{a}^{\mathrm{2}} −{x}} \\ $$$$\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} }{\:{x}}=\frac{{a}^{\mathrm{2}} }{\frac{\mathrm{4}{a}^{\mathrm{2}} {b}}{\mathrm{4}{b}^{\mathrm{2}} \:+\:\mathrm{1}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} ×\frac{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} {b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{b}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{b}+\mathrm{1}}{\mathrm{4}{b}^{\mathrm{2}} −\mathrm{4}{b}+\mathrm{1}}=\frac{\left(\mathrm{2}{b}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{2}{b}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{p}}{{q}}=\frac{\mathrm{2}{b}+\mathrm{1}}{\mathrm{2}{b}−\mathrm{1}}\:\mid\:\frac{{p}}{{q}}=\frac{\mathrm{1}+\mathrm{2}{b}}{\mathrm{1}−\mathrm{2}{b}} \\ $$$$\frac{{p}+{q}}{{p}−{q}}=\frac{\mathrm{4}{b}}{\mathrm{2}}\:=\mathrm{2}{b}\mid\:\frac{{p}+{q}}{{p}−{q}}=\frac{\mathrm{2}}{\mathrm{4}{b}}=\frac{\mathrm{1}}{\mathrm{2}{b}} \\ $$
Answered by som(math1967) last updated on 21/Sep/24
(((√(a^2 +x))+(√(a^2 −x)))/( (√(a^z +x))−(√(a^2 −x)))) =((((√(a^2 +x))+(√(a^2 −x)))^2 )/(((√(a^2 +x)))^2 −((√(a^2 −x)))^2 )) =((a^2 +x+a^2 −x+2(√(a^4 −x^2 )))/(a^2 +x−a^2 +x)) =((2(a^2 +(√(a^4 −x^2 )))/(2x)) =((a^2 +(√(a^4 −((16a^4 b^2 )/((4b^2 +1)^2 )))))/((4a^2 b)/(4b^2 +1))) =((a^2 +a^2 (√(((4b^2 +1)^2 −4.4b^2 .1)/((4b^2 +1)^2 ))))/((4a^2 b)/((4b^2 +1)))) =((a^2 (1+((4b^2 −1)/(4b^2 +1))))/((4a^2 b)/(4b^2 +1))) =((a^2 ×8b^2 )/(4b^2 +1))×((4b^2 +1)/(4a^2 b))=2b
$$\:\frac{\sqrt{{a}^{\mathrm{2}} +{x}}+\sqrt{{a}^{\mathrm{2}} −{x}}}{\:\sqrt{{a}^{{z}} +{x}}−\sqrt{{a}^{\mathrm{2}} −{x}}} \\ $$$$=\frac{\left(\sqrt{{a}^{\mathrm{2}} +{x}}+\sqrt{{a}^{\mathrm{2}} −{x}}\right)^{\mathrm{2}} }{\left(\sqrt{{a}^{\mathrm{2}} +{x}}\right)^{\mathrm{2}} −\left(\sqrt{{a}^{\mathrm{2}} −{x}}\right)^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} +{x}+{a}^{\mathrm{2}} −{x}+\mathrm{2}\sqrt{{a}^{\mathrm{4}} −{x}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{x}−{a}^{\mathrm{2}} +{x}} \\ $$$$=\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{4}} −{x}^{\mathrm{2}} }\right.}{\mathrm{2}{x}} \\ $$$$=\frac{{a}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{4}} −\frac{\mathrm{16}{a}^{\mathrm{4}} {b}^{\mathrm{2}} }{\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }}}{\frac{\mathrm{4}{a}^{\mathrm{2}} {b}}{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\frac{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} \sqrt{\frac{\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{4}{b}^{\mathrm{2}} .\mathrm{1}}{\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }}}{\frac{\mathrm{4}{a}^{\mathrm{2}} {b}}{\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}\right)}} \\ $$$$=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{4}{b}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}\right)}{\frac{\mathrm{4}{a}^{\mathrm{2}} {b}}{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\frac{{a}^{\mathrm{2}} ×\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}×\frac{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} {b}}=\mathrm{2}{b} \\ $$$$ \\ $$
Commented by som(math1967) last updated on 21/Sep/24
yes sir,
$${yes}\:{sir}, \\ $$
Commented by MATHEMATICSAM last updated on 21/Sep/24
sir the answer should be 2b
$$\mathrm{sir}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{2}{b} \\ $$

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