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