Question

simplify √242 + √128-√72​

Answer 1

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Answer 2

$\red{ Value \: of\: \sqrt{242} + \sqrt{128} - \sqrt{72} }$

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$We \: have ,$

$i) \sqrt{242} \\= \sqrt{11\times 11 \times 2 } \\= 11\sqrt{2}$

$ii) \sqrt{128} \\= \sqrt{2^{6} \times 2 } \\= 2^{3} \times \sqrt{2}\\= 8\sqrt{2}$

$iii) \sqrt{72} \\= \sqrt{6\times 6 \times 2 } \\= 6\sqrt{2}$

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$= 11\sqrt{2} + 8\sqrt{2} - 6\sqrt{2} \\= (11 + 8 - 6 )\sqrt{2} \\= ( 19 - 6) \sqrt{2} \\= 13\sqrt{2} \\= 13 \times 1.414 \\\approx 18.382$

Therefore.,

$\red{ Value \: of\: \sqrt{242} + \sqrt{128} - \sqrt{72} } \\\green {= 13\sqrt{2}}\\\approx { 18.382}$

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