Question

5×2√3/7+4√3 =a+b√3 find the value of a and b

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{5 \times 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3} \\ \\ solve \\ on \: rationalizing \: we \: get \\ \\ \frac{10 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = \frac{10 \sqrt{3} \times 7 - 10 \sqrt{3} \times 4 \sqrt{3} }{ {(7)}^{2} - (4 \sqrt{3})^{2}} \\ \\ = \frac{70 \sqrt{3} - 120}{4 9- 48} \\ \\ = 70 \sqrt{3} - 120 \\ \\ - 120 + 70 \sqrt{3} = a + b \sqrt{3} \\ \\ a = - 120 \\ b = 70$

Hope this helps!!!!

@Mahak24

Thanks...
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Answer 2

Hello friends!!

Here is ur answer :

$\frac{5 \times 2 \sqrt{3} } {7 + 4 \sqrt{3} } = a + b \sqrt{3}$


$\frac{10 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$


$\frac{10 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3}$


$\frac{(10 \sqrt{3} )(7 - 4 \sqrt{3} )}{(7 + 4 \sqrt{3})(7 - 4 \sqrt{3}) } = a + b \sqrt{3}$


Using identity :

( a + b) ( a - b) = a² - b²


$\frac{70 \sqrt{3} - 120 }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } = a + b \sqrt{3}$



$\frac{70 \sqrt{3} - 120 }{49 - 48} = a + b \sqrt{3}$


$\frac{70 \sqrt{3} - 120 }{1} = a + b \sqrt{3}$




$- 120 + 70 \sqrt{3} = a + b \sqrt{3}$



Value of

$a = - 120$

$b = 70$



HOPE IT HELPS YOU..