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Question Number 47531 by peter frank last updated on 11/Nov/18

Commented by math1967 last updated on 11/Nov/18

question should be A±(√((A+G)(A−G)))

$${question}\:{should}\:{be}\:{A}\pm\sqrt{\left({A}+{G}\right)\left({A}−{G}\right)} \\ $$

Commented by peter frank last updated on 11/Nov/18

yes sir.thanks for correction

$$\mathrm{yes}\:\mathrm{sir}.\mathrm{thanks}\:\mathrm{for}\:\mathrm{correction} \\ $$

Answered by math1967 last updated on 11/Nov/18

let two numbers are a,b ∴A=((a+b)/2) ,G=(√(ab)) ∴a+b=2A again (a−b)^2 =(a+b)^2 −4ab ⇒(a−b)^2 =4A^2 −4G ∴a−b=±2(√(A^2 −G^2 )) a+b=2A by solving two numbers are A+(√(A^2 −G^2 )) and A−(√(A^2 −G^2 )) or A±(√((A+G)(A−G)))

$${let}\:{two}\:{numbers}\:{are}\:{a},{b} \\ $$$$\therefore{A}=\frac{{a}+{b}}{\mathrm{2}}\:,{G}=\sqrt{{ab}}\: \\ $$$$\:\:\therefore{a}+{b}=\mathrm{2}{A}\: \\ $$$${again}\:\left({a}−{b}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}{ab} \\ $$$$\:\:\Rightarrow\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{4}{A}^{\mathrm{2}} −\mathrm{4}{G} \\ $$$$\:\therefore{a}−{b}=\pm\mathrm{2}\sqrt{{A}^{\mathrm{2}} −{G}^{\mathrm{2}} } \\ $$$$\:\:\:{a}+{b}=\mathrm{2}{A} \\ $$$${by}\:{solving}\:{two}\:{numbers}\:{are} \\ $$$${A}+\sqrt{{A}^{\mathrm{2}} −{G}^{\mathrm{2}} \:\:}\:{and}\:{A}−\sqrt{{A}^{\mathrm{2}} −{G}^{\mathrm{2}} } \\ $$$${or}\:{A}\pm\sqrt{\left({A}+{G}\right)\left({A}−{G}\right)} \\ $$

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