Question

3+underoot7 over 3-4 underoot7 =a+bunderoot7​

Answer 1

Question :-

$\dfrac{3+\sqrt{7}}{3-4\sqrt{7}} = a + \sqrt{b}$

◾Solution :-

⏩As we are able to see the denominator is irrational , So we will first rationalize the denominator.

◾Rationalizing :-

$= \dfrac{3+\sqrt{7}}{3-4\sqrt{7}} \times \dfrac{3+4\sqrt{7}}{3+4\sqrt{7}}$

$= \dfrac{(3+\sqrt{7})(3+4\sqrt{7})}{(3)^2-(4\sqrt{7})^2}$

$= \dfrac{ 9 + 12\sqrt{7} + 3\sqrt{7} + 28 }{9 - 112}$

$= \dfrac{ 37 + 15\sqrt{7}}{-103}$

$= \dfrac{-37 - 15\sqrt{7}}{103}$

◾So the values of

$\bold{ a = \dfrac{-37}{103} }$

$\bold{b = \dfrac{-15}{103}}$

Answer 2

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

Rationalising Factor of 3-4√7=3+4√7

Multiply Both Numerator and Denominator with 3+4√7

$\frac{(3 + \sqrt{7}) (3 + 4 \sqrt{7}) }{(3 - 4 \sqrt{7})(3 + 4 \sqrt{7}) } = a + b \sqrt{7} \\ \\ \frac{3(3 + 4 \sqrt{7} ) + \sqrt{7}(3 + 4 \sqrt{7} )}{ {3}^{2} -( 4 \sqrt{7)} {}^{2} } \\ \\ \frac{9 + 12 \sqrt{7} + 3 \sqrt{7} + 4 \times 7}{9 - 16 \times 7} \\ \\ \frac{9 + 12 \sqrt{7} + 3 \sqrt{7} + 28 }{9 - 112} \\ \\ \frac{37 + 15 \sqrt{7} }{ - 103} \\ \\ \frac{ - (37 + 15 \sqrt{7} )}{103} \\ \\ \frac{ - 37 - 15 \sqrt{7} }{103} = a + b \sqrt{7} \\ \\ - \frac{ 37}{103} - \frac{15}{103} \times \sqrt{7}$

Comparing Both Sides we get

a=-37/103

b=-15/103