Question

simplify ( 4+√3) (4-√3)​

Answer 1

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

$\frac{4 + \sqrt{3} }{4 - \sqrt{3} }$

${\blue{\underline{\underline{\bold{Concept\:used\:here-}}}}}$

• Rationalizing the denominator.

${\green{\underline{\underline{\bold{Solution:-}}}}}$

$\frac{4 + \sqrt{3} }{4 - \sqrt{3} }$

By rationalizing the denominator:-

$\frac{4 + \sqrt{3} }{4 - \sqrt{3} } \times \frac{4 + \sqrt{3} }{4 + \sqrt{3} } \\ \\ = \frac{ {(4 + \sqrt{3} )}^{2} }{(4 - \sqrt{3} )(4 + \sqrt{3} )} \: \: \: ( \therefore(a + b)(a - b) = {a}^{2} - {b}^{2} ) \\ \\ = \frac{ {(4 + \sqrt{3}) }^{2} }{ {4}^{2} - {( \sqrt{3} )}^{2} } \: \: \: ( \therefore( { a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} ) \\ \\ = \frac{ {4}^{2} + { (\sqrt{3})}^{2} + 2 \times 4 \times \sqrt{3} }{4 - 3} \\ \\ = \frac{16 + 3 + 8 \sqrt{3} }{1} \\ \\ = 19 + 8\sqrt{3}$

Therefore :-

$\frac{4 + \sqrt{3} }{4 - \sqrt{3} }$

= $\boxed{19 + 8 \sqrt{3} }$