Question

Find the value of a and b in each of the following equalities : 5 + 2√3 / 7+ 4√3 = a + b √3

Answer 1

$\large\bf\underline{Given:-}$

• 5+2√3 /7+4√3 = a+ b√3

$\large\bf\underline {To \: find:-}$

• Value of a and b.

$\huge\bf\underline{Solution:-}$

• 5 + 2√3 / 7+ 4√3 = a + b √3

$\rm \dashrightarrow \: \frac{5 + 2 \sqrt{3} }{ 7 + 4 \sqrt{3} } = a + b \sqrt{3} \\ \\ \rm \dashrightarrow \:taking \: \underline{LHS : - } \\ \\ \rm\rm \dashrightarrow \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ \rm \underline{ \dag \: rationalising \: denominator : \: - } \\ \\ \rm \dashrightarrow \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ \bf \dashrightarrow \: (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ \rm \dashrightarrow \frac{5 (7 - 4 \sqrt{3}) + 2 \sqrt{3}(7 - 4 \sqrt{3} ) }{49 - 48} \\ \\ \rm \dashrightarrow \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{1} \\ \\\rm{\text{it is given that LHS = a + b root3}} \\\\ \rm \dashrightarrow \frac{11 - 6 \sqrt{3} }{1} = a + b \sqrt{3} \\ \\ \dashrightarrow \boxed{ \bf 11 = a} \\ \\ \rm \dashrightarrow \: b \sqrt{3} = - 6 \sqrt{3} \\ \\ \rm \dashrightarrow \boxed{ \bf \: b = - 6}$

✞ Hence , a = 11 and b = -6

Answer 2

$\bf\large{\underline{Question:-}}$

Find the value of a and b in each of the following equalities : 5 + 2√3 / 7+ 4√3 = a + b √3.

$\bf\large{\underline{Given:-}}$

• 5 + 2√3 / 7+ 4√3 .

$\bf\large{\underline{To\:find:-}}$

• value of a+b=?

$\bf\large{\underline{Solution:-}}$

Identity used

$\tt→ a^2-b^2=(a+b)(a-b)$

Now,

★ Rationalize the denominator

$\tt→ \frac{5+2\sqrt3}{7+4\sqrt3}×\frac{7-4\sqrt3}{7-4\sqrt3}=a+b\sqrt3\\\tt→5 \frac{(7-4\sqrt3)+2\sqrt3(7-4\sqrt3)}{49-48}=a+b\sqrt3\\\tt→ \frac{35-20\sqrt3+14\sqrt3-24}{49-48}=a+b\sqrt3\\\tt→ \frac{11-6\sqrt3}{1}=a+b\sqrt3\\\tt→ a=11 \\\tt→ b\sqrt3=6\sqrt3 (\sqrt3\:,\sqrt3 \: cancel\:out\:by\:each\:other)\\\tt→ b=-6$

Hence,

→ a= 11

→ b= -6