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Question Number 191947 by universe last updated on 04/May/23
prove that  (√(a+b(√(a−b(√(a+b(√(a−b(√(...))))))))))  =  (((√(4a−3b^2 ))+b)/2)
$${prove}\:{that} \\ $$$$\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{…}}}}}\:\:=\:\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }+{b}}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 05/May/23
Let s=(√(a+b(√(a−bs))))=((b+(√(4a−3b^2 )))/2)  ⇒ s^2 =a+b(√(a−bs))  (s^2 −a)^2 =b^2 (a−bs)   ...(i)  &    (2s−b)^2 =4a−3b^2       ⇒  4s^2 +b^2 −4bs=4a−3b^2   or       (s^2 −a)^2 =b^2 (s−b)^2   ..(ii)  comparing (i) & (ii)  ⇒  (s−b)^2 =a−bs  (s−b)^2 +b(s−b)−(a−b^2 )=0  s−b=((−b±(√(4a−3b^2 )))/2)  s=((b±(√(4a−3b^2 )))/2)
$${Let}\:{s}=\sqrt{{a}+{b}\sqrt{{a}−{bs}}}=\frac{{b}+\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\:{s}^{\mathrm{2}} ={a}+{b}\sqrt{{a}−{bs}} \\ $$$$\left({s}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \left({a}−{bs}\right)\:\:\:…\left({i}\right) \\ $$$$\&\:\:\:\:\left(\mathrm{2}{s}−{b}\right)^{\mathrm{2}} =\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} \:\:\:\: \\ $$$$\Rightarrow\:\:\mathrm{4}{s}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{4}{bs}=\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} \\ $$$${or}\:\:\:\:\:\:\:\left({s}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \left({s}−{b}\right)^{\mathrm{2}} \:\:..\left({ii}\right) \\ $$$${comparing}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\Rightarrow\:\:\left({s}−{b}\right)^{\mathrm{2}} ={a}−{bs} \\ $$$$\left({s}−{b}\right)^{\mathrm{2}} +{b}\left({s}−{b}\right)−\left({a}−{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${s}−{b}=\frac{−{b}\pm\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${s}=\frac{{b}\pm\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by mr W last updated on 04/May/23
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Commented by mehdee42 last updated on 05/May/23
dear friend,what is the reason for using equation (2s−b)^2 =4a−3b^2   ?
$${dear}\:{friend},{what}\:{is}\:{the}\:{reason}\:{for}\:{using}\:{equation}\:\left(\mathrm{2}{s}−{b}\right)^{\mathrm{2}} =\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} \:\:? \\ $$
Answered by universe last updated on 05/May/23
   (√(a+b(√(a−b(√(a+b(√(a−b(√(...)))))))))) = x (let)  ...(1)     and     (√(a−b(√(a+b(√(a−b(√(a+b(√(...))))))))))  =  y(let) ...(2)     by (1)      (√(a+by))  =  x  ⇒ a+by = x^2   ....(3)    now by (2)     (√(a−bx))  =  y  ⇒ a−bx  = y^2     ....(4)    eq^n (3)−eq^n (4)     b(x+y)  =   (x−y)(x+y) ⇒ x−y = b     y  = x−b     by (3)     a+bx−b^2  = x^2   ⇒ x^2 −bx +b^2 −a = 0     x   =  ((b+(√(b^2 −4(b^2 −a))))/2)     x =  ((b+(√(4a−3b^2 )))/2)
$$\:\:\:\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{…}}}}}\:=\:{x}\:\left({let}\right)\:\:…\left(\mathrm{1}\right) \\ $$$$\:\:\:{and} \\ $$$$\:\:\:\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{…}}}}}\:\:=\:\:{y}\left({let}\right)\:…\left(\mathrm{2}\right) \\ $$$$\:\:\:{by}\:\left(\mathrm{1}\right)\: \\ $$$$\:\:\:\sqrt{{a}+{by}}\:\:=\:\:{x}\:\:\Rightarrow\:{a}+{by}\:=\:{x}^{\mathrm{2}} \:\:….\left(\mathrm{3}\right) \\ $$$$\:\:{now}\:{by}\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\sqrt{{a}−{bx}}\:\:=\:\:{y}\:\:\Rightarrow\:{a}−{bx}\:\:=\:{y}^{\mathrm{2}} \:\:\:\:….\left(\mathrm{4}\right) \\ $$$$\:\:{eq}^{{n}} \left(\mathrm{3}\right)−{eq}^{{n}} \left(\mathrm{4}\right) \\ $$$$\:\:\:{b}\left({x}+{y}\right)\:\:=\:\:\:\left({x}−{y}\right)\left({x}+{y}\right)\:\Rightarrow\:{x}−{y}\:=\:{b} \\ $$$$\:\:\:{y}\:\:=\:{x}−{b} \\ $$$$\:\:\:{by}\:\left(\mathrm{3}\right) \\ $$$$\:\:\:{a}+{bx}−{b}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \:\:\Rightarrow\:{x}^{\mathrm{2}} −{bx}\:+{b}^{\mathrm{2}} −{a}\:=\:\mathrm{0} \\ $$$$\:\:\:{x}\:\:\:=\:\:\frac{{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −{a}\right)}}{\mathrm{2}} \\ $$$$\:\:\:{x}\:=\:\:\frac{{b}+\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by mehdee42 last updated on 05/May/23
a beautiful solution
$${a}\:{beautiful}\:{solution}\: \\ $$

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