Question

5+3 root 3/7+4 root 3=a+b root 3
Plz solve this quickly

Answer 1

$\bf \large \it{Hey \: User!!!}$

$\bf \: \: given = > \frac{5 + 3 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3} \\ \\ LHS :- \\ \\ > > \frac{5 + 3 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ > > \frac{(5 + 3 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 + 4\sqrt{3} )(7 - 4 \sqrt{3} )} \\ \\ > > \frac{5(7 - 4 \sqrt{3} ) + 3 \sqrt{3}(7 - 4 \sqrt{3} ) }{( {7})^{2} - ( {4 \sqrt{3} )}^{2} } \\ \\ > > \frac{35 - 20 \sqrt{ 3} + 7 \sqrt{3} - 36}{49 - 48} \\ \\ > > \frac{ - 1 - 13 \sqrt{3} }{1} = - 1 - 13\sqrt{3} \\ \\ RHS :- \\ \\ \bf = a + b \sqrt{3} \\ \\ \bf \: on \: comparing \: LHS \: with \: RHS \\ \\ \bf \: > > - 1 - 13 \sqrt{3} = a + b \sqrt{3} \\ \\ \bf \: therefore \: a = - 1 \: and \: b = - 13 \\ \\ \bf \: verification : - \\ \\ > > - 1 + ( - 13) \sqrt{3} = - 1 - 13 \sqrt{3}$

$\bf \large \it \: Cheers!!! \:$

Answer 2

Answer:

a = -1 and b = 1

Step-by-step explanation:

We have,

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

For rationalizing R.H.S.,

Multiply and divide 7 - 4√3 on the right side,

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

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

$\frac{(5+3\sqrt{3})(7-4\sqrt{3})}{49-48}=a+b\sqrt{3}$

$\frac{35+21\sqrt{3}-20\sqrt{3}-36}{1}=a+b\sqrt{3}$

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

By comparing both sides,

We get, a = -1 and b = 1