Find the value of a and b if 7 + 4 √ 3 ÷ 7 - 2√ 3 = a - √ 3 b

1.	Find the value of a and b if 7 + 4 √ 3 ÷ 7 - 2√ 3 = a - √ 3 b
Answer» ANSWER: Given: $\sf{(7 + 4\sqrt3) \div (7-2\sqrt3) = a-b\sqrt3}$ To FIND: VALUES of a and b Solution: $:\longrightarrow \sf{(7 + 4\sqrt3) \div (7-2\sqrt3) = a-b\sqrt3} \\\\:\implies \sf{\dfrac{7+4\sqrt3}{7-2\sqrt3} = a-b\sqrt3} \\\\\text{Rationalising the denominator on LHS,} \\\\:\implies \sf{\dfrac{7+4\sqrt3}{7-2\sqrt3} \times \dfrac{7+2\sqrt3}{7+2\sqrt3} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{(7+4\sqrt3)\times(7+2\sqrt3)}{(7-2\sqrt3)\times(7+2\sqrt3)} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{(49+14\sqrt3+28\sqrt3+8(\sqrt3)^2)}{(7^2 - (2\sqrt3)^2)} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{(49+24+42\sqrt3)}{(49-12)} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{73+42\sqrt3}{37} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{73}{37} + \dfrac{42\sqrt3}{37} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{73}{37} - \dfrac{(-42\sqrt3)}{37} = a-b\sqrt3} \\\\\text{Comparing the terms, in LHS and RHS,} \\\\\bf{:\implies a = \dfrac{73}{37}\: \& \:b = \dfrac{-42}{37}}$ Formula Used: (a + b)(a - b) = a² - b² Learn More: $\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$

Answer» ANSWER:

Given:

$\sf{(7 + 4\sqrt3) \div (7-2\sqrt3) = a-b\sqrt3}$

To FIND:

VALUES of a and b

Solution:

$:\longrightarrow \sf{(7 + 4\sqrt3) \div (7-2\sqrt3) = a-b\sqrt3} \\\\:\implies \sf{\dfrac{7+4\sqrt3}{7-2\sqrt3} = a-b\sqrt3} \\\\\text{Rationalising the denominator on LHS,} \\\\:\implies \sf{\dfrac{7+4\sqrt3}{7-2\sqrt3} \times \dfrac{7+2\sqrt3}{7+2\sqrt3} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{(7+4\sqrt3)\times(7+2\sqrt3)}{(7-2\sqrt3)\times(7+2\sqrt3)} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{(49+14\sqrt3+28\sqrt3+8(\sqrt3)^2)}{(7^2 - (2\sqrt3)^2)} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{(49+24+42\sqrt3)}{(49-12)} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{73+42\sqrt3}{37} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{73}{37} + \dfrac{42\sqrt3}{37} = a-b\sqrt3} \\\\:\implies \sf{\dfrac{73}{37} - \dfrac{(-42\sqrt3)}{37} = a-b\sqrt3} \\\\\text{Comparing the terms, in LHS and RHS,} \\\\\bf{:\implies a = \dfrac{73}{37}\: \& \:b = \dfrac{-42}{37}}$

Formula Used:

(a + b)(a - b) = a² - b²

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$