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If $a \sqrt{3} + + b \sqrt{3} = \sqrt{75}$ , where , then the number of possible ordered pairs of the form $(a, b)$ can be obtained is

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Detailed Solution

$a \sqrt{3} + + b \sqrt{3} = \sqrt{75}$

$a \sqrt{3} + + b \sqrt{3} = 5 \sqrt{3}$

$(a + b) \sqrt{3} = 5 \sqrt{3}$

$a + b = 5$

Possible ordered pairs ${(14), (2, 3), (3, 2), (4, 1)]$

No., of possible ordered pairs is 4.

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