If $(a + b)^{2} = 4 a b$ , then prove that $a = b$ .

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Solution

${(a + b)}^{2} = 4 a b$
$a^{2} + b^{2} + 2 a b = 4 a b$

$a^{2} + b^{2} - 2 a b = 0$

${(a - b)}^{2} = 0$

$a = b$
Hence, proved.

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Similar questions

{(a + b)}^{2} - (a - b)^{2} = 4 a b

(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})

(a + b)^{2} - (a - b)^{2} = 4 a b

b^{2} = 4 a b

x

\sqrt{a + x} + \sqrt{a - x} = \sqrt{b}

x = \frac{4 a b}{a + b}

\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = 2

B = - A^{- 1} B A

(A + B)^{2} = A^{2} + B^{2}

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