Question

if
$\sqrt{3} ( \sqrt{7} - \sqrt{3} ) = \sqrt{a } + b \:$
what is the value of a and b​

Answer 1

Step-by-step explanation:

$\huge\tt\underline\blue{⛶Answer⛶ }$

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⟹$\frac{8 + 3 \sqrt{7} }{8 - 3 \sqrt{7} } - \frac{8 - 3 \sqrt{7} }{8 + 3 \sqrt{7} } = a + b \sqrt{7}$

⟹ $\frac{(8 + 3 \sqrt{7})(8 + 3 \sqrt{7}) - [(8 - 3 \sqrt{7} )(8 - 3 \sqrt{7}) ]}{(8 + 3 \sqrt{7} )(8 - 3 \sqrt{7} )}$

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

⟹ $\frac{4 \times 8 \sqrt{7} }{ {(8)}^{2} - {(3 \sqrt{7}) }^{2} } = a + b \sqrt{7}$

⟹$\frac{32 \sqrt{7} }{64 - 63} = a + b \sqrt{7}$

⟹ $\frac{32 \sqrt{7} }{1} = a + b \sqrt{7}$

⟹$32 \sqrt{7} = a + b \sqrt{7}⟹32$

⟹0 + 32 $\sqrt{7} = a + b \sqrt{7}⟹0+32$

On comparing both sides:-

⟹a = 0 $\: and \: b = 32⟹a=0andb=32$

⟹now $\: \sqrt{a - b} = \sqrt{0 - 32} = \sqrt{ - 32 } = \sqrt{ - 2 \times 16} = 4 \sqrt{ - 2} = 4 \sqrt{2} i \: [( {i}^{2} = - 1)$

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Answer 2

Answer:

thanks for free point............