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Question Number 149889 by Samimsultani last updated on 08/Aug/21

Commented by bramlexs22 last updated on 08/Aug/21

(√(a+b(√d))) = (√((a+c)/2)) +sgn(b)(√((a−c)/2)) where c=(√(a^2 −b^2 d))

$$\sqrt{\mathrm{a}+\mathrm{b}\sqrt{\mathrm{d}}}\:=\:\sqrt{\frac{\mathrm{a}+\mathrm{c}}{\mathrm{2}}}\:+\mathrm{sgn}\left(\mathrm{b}\right)\sqrt{\frac{\mathrm{a}−\mathrm{c}}{\mathrm{2}}} \\ $$$$\mathrm{where}\:\mathrm{c}=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{d}}\: \\ $$

Commented by Samimsultani last updated on 08/Aug/21

please proof???

$$ \\ $$$$\:\:{please}\:{proof}??? \\ $$

Commented by bramlexs22 last updated on 08/Aug/21

(√(a+b(√d))) = (√x) +(√y) ⇒a+b(√d) = x+y+2(√(xy )) → { ((a=x+y⇒x=a−y)),(((√(b^2 d)) =(√(4xy)) ⇒b^2 d=4xy⇒y=((b^2 d)/(4x)))) :} ⇔x=a−((b^2 d)/(4x)) ⇔4x^2 =4ax−b^2 d ⇔4x^2 −4ax+b^2 d ⇔ x = ((4a ± (√(16a^2 −16b^2 d)))/8) ⇒ x=((a ±(√(a^2 −b^2 d)) )/2) ; let c=(√(a^2 −b^2 d)) ⇒ { ((x_1 =((a+c)/2))),((x_2 =((a−c)/2))) :}

$$\sqrt{\mathrm{a}+\mathrm{b}\sqrt{\mathrm{d}}}\:=\:\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{y}} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}\sqrt{\mathrm{d}}\:=\:\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}\:} \\ $$$$\rightarrow\begin{cases}{\mathrm{a}=\mathrm{x}+\mathrm{y}\Rightarrow\mathrm{x}=\mathrm{a}−\mathrm{y}}\\{\sqrt{\mathrm{b}^{\mathrm{2}} \mathrm{d}}\:=\sqrt{\mathrm{4xy}}\:\Rightarrow\mathrm{b}^{\mathrm{2}} \mathrm{d}=\mathrm{4xy}\Rightarrow\mathrm{y}=\frac{\mathrm{b}^{\mathrm{2}} \mathrm{d}}{\mathrm{4x}}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{x}=\mathrm{a}−\frac{\mathrm{b}^{\mathrm{2}} \mathrm{d}}{\mathrm{4x}} \\ $$$$\Leftrightarrow\mathrm{4x}^{\mathrm{2}} =\mathrm{4ax}−\mathrm{b}^{\mathrm{2}} \mathrm{d} \\ $$$$\Leftrightarrow\mathrm{4x}^{\mathrm{2}} −\mathrm{4ax}+\mathrm{b}^{\mathrm{2}} \mathrm{d}\: \\ $$$$\Leftrightarrow\:\mathrm{x}\:=\:\frac{\mathrm{4a}\:\pm\:\sqrt{\mathrm{16a}^{\mathrm{2}} −\mathrm{16b}^{\mathrm{2}} \mathrm{d}}}{\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{x}=\frac{\mathrm{a}\:\pm\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{d}}\:}{\mathrm{2}}\:;\:\mathrm{let}\:\mathrm{c}=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{d}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{a}+\mathrm{c}}{\mathrm{2}}}\\{\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{a}−\mathrm{c}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$

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