Question

Plase answer this math question

Answer 1

AnsWer :

11 + ( - 6 √3 )

SolutioN :

$\tt \dagger \: \: \: \: \: \dfrac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$

$\tt \rightarrow \dfrac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\tt \rightarrow \dfrac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \dfrac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

$\tt \rightarrow \dfrac{(5 + 2 \sqrt{3}) \times( 7 - 4 \sqrt{3})}{{(7)}^{2} - {( 4 \sqrt{3})}^{2}}$

$\tt \rightarrow \dfrac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{{(7)}^{2} - {( 4 \sqrt{3})}^{2}}$

$\tt \rightarrow \dfrac{35 - 6 \sqrt{3} - 24}{{(7)}^{2} - {( 4 \sqrt{3})}^{2}}$

$\tt \rightarrow\dfrac{11 - 6 \sqrt{3} }{{(7)}^{2} - {( 4 \sqrt{3})}^{2}}$

$\tt \rightarrow\dfrac{11 - 6 \sqrt{3} }{49 - {48}}$

$\tt \rightarrow11 - 6 \sqrt{3}$

So,We can also write as,

$\rightarrow11 - 6 \sqrt{3} = a + b \sqrt{3}$

$\rightarrow11 + ( - 6 \sqrt{3} )= a + b \sqrt{3}$

$\rule{100}2$

$\rightarrow a = 11.$

$\rightarrow b \sqrt{3} = - 6 \sqrt{3}$

Therefore, the value of a = 11 and b = - 6.

Answer 2

$\sf{\underline{\large{\underline{\orange{QUESTION}:-}}}}$

Simplify and find the value of “a” and “b” .

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

$\sf{\underline{\large{\underline{\green{ANSWER}:-}}}}$

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3} \\ for \: rationalise \: we \: use \: the \: formula \\ \: \: \: \: (a + b)(a - b) = ( {a}^{2} - {b}^{2} )$

☞Now, first simplify the L.H.S .

✔️L.H.S:-

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ = \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ = \frac{(5 + 2 \sqrt{3})(7 - 4 \sqrt{3}) }{{7}^{2} - {(4 \sqrt{3}) }^{2} } \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{49 - 48} \\ = \frac{11 - 6 \sqrt{3} }{1} \\ = 11 - 6 \sqrt{3}$

☞Now, we see the similarity between L.H.S and R.H.S .

✔️=> 11 - 6√3 = a + b√3

✔️=> 11 + (-6)√3 = a + b√3

☞Now, compare L.H.S and R.H.S we get,

• “a = 11” and “b = -6”

$\bigstar\:\underline{\boxed{\bf{\red{Required\: Answer:\:a\:=\:11\:and\:b\:=\:-6}}}}$