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Question Number 192129 by universe last updated on 08/May/23
prove that ∣a+(√(a^2 −b^2 ))∣ + ∣a − (√(a^2 −b^2 ))∣ = ∣a+b∣ +∣a−b∣ a,b ∈ C
$$\:{prove}\:{that} \\ $$$$\:\mid{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid\:+\:\mid{a}\:−\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid\:=\:\mid{a}+{b}\mid\:+\mid{a}−{b}\mid \\ $$$${a},{b}\:\in\:\mathbb{C} \\ $$
Answered by AST last updated on 08/May/23
Squaring both sides (∣x∣^2 =xx^− ;∣xy∣=∣x∣∣y∣;x+y^(_____) =x^− +y^− ) LHS^2 = (a+(√(a^2 −b^2 )))(a^− +(√(a^2 −b^2 ))^() )+(a−(√(a^2 −b^2 )))(a^− −(√(a^2 −b^2 ))^() ) +2∣(a+(√(a^2 −b^2 )))(a−(√(a^2 −b^2 )))=b^2 ∣ =2∣a∣^2 +2∣a^2 −b^2 ∣+2∣b∣^2 RHS^2 =(a+b)(a^− +b^− )+(a−b)(a^− −b^− )+2∣a^2 −b^2 ∣ =2∣a∣^2 +2∣a^2 −b^2 ∣+2∣b∣^2 Since LHS and RHS were both positive before squaring both sides LHS^2 =RHS^2 ⇒LHS=RHS □
$${Squaring}\:{both}\:{sides} \\ $$$$\left(\mid{x}\mid^{\mathrm{2}} ={x}\overset{−} {{x}};\mid{xy}\mid=\mid{x}\mid\mid{y}\mid;\overset{\_\_\_\_\_} {{x}+{y}}=\overset{−} {{x}}+\overset{−} {{y}}\right) \\ $$$${LHS}^{\mathrm{2}} = \\ $$$$\left({a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\left(\overset{−} {{a}}+\overset{} {\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right)+\left({a}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\left(\overset{−} {{a}}−\overset{} {\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$$+\mathrm{2}\mid\left({a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\left({a}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)={b}^{\mathrm{2}} \mid \\ $$$$=\mathrm{2}\mid{a}\mid^{\mathrm{2}} +\mathrm{2}\mid{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \mid+\mathrm{2}\mid{b}\mid^{\mathrm{2}} \\ $$$${RHS}^{\mathrm{2}} =\left({a}+{b}\right)\left(\overset{−} {{a}}+\overset{−} {{b}}\right)+\left({a}−{b}\right)\left(\overset{−} {{a}}−\overset{−} {{b}}\right)+\mathrm{2}\mid{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \mid \\ $$$$=\mathrm{2}\mid{a}\mid^{\mathrm{2}} +\mathrm{2}\mid{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \mid+\mathrm{2}\mid{b}\mid^{\mathrm{2}} \\ $$$${Since}\:{LHS}\:{and}\:{RHS}\:{were}\:{both}\:{positive}\:{before} \\ $$$${squaring}\:{both}\:{sides} \\ $$$${LHS}^{\mathrm{2}} ={RHS}^{\mathrm{2}} \Rightarrow{LHS}={RHS}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\ $$
Commented by York12 last updated on 08/May/23
sir I wanna ask you[several quesions , I am a high school student and I wanna ask about books recommendationd so sir that is my telegram : bengubler
$${sir}\:{I}\:{wanna}\:{ask}\:{you}\left[{several}\:{quesions}\:,\right. \\ $$$$\:{I}\:{am}\:{a}\:{high}\:{school}\:{student}\: \\ $$$${and}\:{I}\:{wanna}\:{ask}\:{about}\:{books}\:{recommendationd} \\ $$$${so}\:{sir}\:{that}\:{is}\:{my}\:{telegram}\:\::\:{bengubler} \\ $$
Commented by AST last updated on 09/May/23
I don′t use telegram.
$${I}\:{don}'{t}\:{use}\:{telegram}. \\ $$
Commented by York12 last updated on 09/May/23
so sir how can I reach you out
$${so}\:{sir}\:{how}\:{can}\:{I}\:{reach}\:{you}\:{out} \\ $$
Answered by universe last updated on 09/May/23
(1/2){∣a+b+a−b+2(√((a+b)(a−b)))∣+∣a+b+a−b−2(√((a+b)(a−b)))∣} let a+b = x and a−b = y (1/2){∣((√x))^2 +((√y))^2 +2(√(xy))∣+∣((√x))^2 +((√y))^2 −2(√(xy))∣} (1/2){∣(√x)+(√y) ∣^2 +∣(√x)− (√y) ∣^2 } let (√x) = u and (√y) = v (1/2){(u+v)(u^� +v^� )+(u−v)(u^� −v^� )} (1/2){uu^� +uv^� +vu^� +vv^� +uu^� −uv^� −vu^� +vv^� } uu^� +vv^� ⇒∣u^2 ∣+∣v^2 ∣ ⇒∣((√x))^2 ∣+∣((√y))^2 ∣ ∣x∣+∣y∣ ⇒ ∣a+b∣ + ∣a−b∣
$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\mid\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{a}}−\boldsymbol{{b}}+\mathrm{2}\sqrt{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)}\mid+\mid\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{a}}−\boldsymbol{{b}}−\mathrm{2}\sqrt{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)}\mid\right\} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{a}}+\boldsymbol{{b}}\:=\:\boldsymbol{{x}} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{a}}−\boldsymbol{{b}}\:=\:\boldsymbol{{y}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\mid\left(\sqrt{\boldsymbol{{x}}}\right)^{\mathrm{2}} +\left(\sqrt{\boldsymbol{{y}}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\boldsymbol{{xy}}}\mid+\mid\left(\sqrt{\boldsymbol{{x}}}\right)^{\mathrm{2}} +\left(\sqrt{\boldsymbol{{y}}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\boldsymbol{{xy}}}\mid\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\mid\sqrt{\boldsymbol{{x}}}+\sqrt{\boldsymbol{{y}}}\:\mid^{\mathrm{2}} +\mid\sqrt{\boldsymbol{{x}}}−\:\sqrt{\boldsymbol{{y}}}\:\mid^{\mathrm{2}} \right\} \\ $$$$\boldsymbol{{let}}\:\sqrt{\boldsymbol{{x}}}\:\:=\:\:\boldsymbol{{u}}\:\:\boldsymbol{{and}}\:\:\sqrt{\boldsymbol{{y}}}\:\:=\:\:\boldsymbol{{v}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\boldsymbol{{u}}+\boldsymbol{{v}}\right)\left(\bar {\boldsymbol{{u}}}+\bar {\boldsymbol{{v}}}\right)+\left(\boldsymbol{{u}}−\boldsymbol{{v}}\right)\left(\bar {\boldsymbol{{u}}}−\bar {\boldsymbol{{v}}}\right)\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\boldsymbol{{u}}\bar {\boldsymbol{{u}}}+\boldsymbol{{u}}\bar {\boldsymbol{{v}}}+\boldsymbol{{v}}\bar {\boldsymbol{{u}}}+\boldsymbol{{v}}\bar {\boldsymbol{{v}}}+\boldsymbol{{u}}\bar {\boldsymbol{{u}}}−\boldsymbol{{u}}\bar {\boldsymbol{{v}}}−\boldsymbol{{v}}\bar {\boldsymbol{{u}}}+\boldsymbol{{v}}\bar {\boldsymbol{{v}}}\right\} \\ $$$$\boldsymbol{{u}}\bar {\boldsymbol{{u}}}+\boldsymbol{{v}}\bar {\boldsymbol{{v}}}\:\Rightarrow\mid\boldsymbol{{u}}^{\mathrm{2}} \mid+\mid\boldsymbol{{v}}^{\mathrm{2}} \mid\:\Rightarrow\mid\left(\sqrt{\boldsymbol{{x}}}\right)^{\mathrm{2}} \mid+\mid\left(\sqrt{\boldsymbol{{y}}}\right)^{\mathrm{2}} \mid \\ $$$$\:\:\mid\boldsymbol{{x}}\mid+\mid\boldsymbol{{y}}\mid\:\Rightarrow\:\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid\:+\:\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$

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