Question

If a be the am and g be the gm of two positive numbers show that numbers are a+-√a^2-g^2

Answer 1

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Answer 2

$\green{ \mathtt{\huge \underline{\fbox{Solution :</p><p> \: \: }}}}$

Given ,

A and G are AM and GM between two positive numbers

Therefore ,

$\large{ \purple{ \star }}$ A = a + b/2

2A = a + b ---- (i)

$\large{ \purple{ \star }}$ G = √ab

G² = ab ---- (ii)

$\star$Put the values of eq (i) and eq (ii) in identity

$\mathtt{\fbox{ {(a - b)}^{2} = {(a + b)}^{2} - 4ab}}$

Thus ,

$\sf \hookrightarrow {(a + b)}^{2} = { (2A)}^{2} - 4 {(G)}^{2} \\ \\ \sf \hookrightarrow {(a + b)}^{2} =4 {(A)}^{2} - 4 {(G )}^{2}$

$\star$Taking square root on both side , we get

$\sf \hookrightarrow a + b= \sqrt{4 {(A)}^{2} - 4 {(G )}^{2} } \\ \\ \sf \hookrightarrow a + b= 2 \sqrt{{(A)}^{2} - {(G )}^{2} } \\ \\ \sf \hookrightarrow a + b = 2 \sqrt{(A + G)(A - G)} \: \: - - - \: \: eq(iii)$

$\star$Add equation i and iii , we get

$\sf \hookrightarrow (a + b) + (a - b) = 2A + 2 \sqrt{(A + G)(A - G)} \\ \\ \sf \hookrightarrow 2a = 2(A + \sqrt{(A + G)(A - G)} ) \\ \\ \sf \hookrightarrow a = A + \sqrt{(A + G)(A - G)}$

$\star$ Put the value of $\sf a = A + \sqrt{(A + G)(A - G)}$ in equation (iii)

$\sf \hookrightarrow A + \sqrt{(A + G)(A - G)} + b = 2A \\ \\ \sf \hookrightarrow b = 2A - ( A + \sqrt{(A + G)(A - G)} ) \\ \\ \sf \hookrightarrow b = 2A - A - \sqrt{(A + G)(A - G)} ) \\ \\ \sf \hookrightarrow b = A - \sqrt{(A + G)(A - G)} )$

Hence , the both numbers are $\sf A ± \sqrt{(A + G)(A - G)} )$