Question

please answer my question as soon as possible because it is very important for me​

Answer 1

a) Solution

→We Have

$\implies\sf \dfrac{\sqrt{3} -1}{\sqrt{3}+1} =a + b\sqrt3$

→Using Rationalization Method

$\implies\sf \dfrac{\sqrt{3} -1}{\sqrt{3}+1} \times\dfrac{\sqrt{3} -1}{\sqrt{3}-1}$

→using this identities

$\sf\implies ( a - b )^2 = a^2+b^2 - 2ab$

$\sf\implies( a - b)(a + b) = a^2- b^2$

→We get

$\sf\implies \dfrac{(\sqrt{3}-1)^2 }{(\sqrt{3} )^2-1}$

$\sf\implies \dfrac{(\sqrt{3})^2+1-2\sqrt{3}}{3-1}$

$\sf\implies \dfrac{3+1-2\sqrt{3}}{2} =\dfrac{4-2\sqrt{3}}{2} =\dfrac{2(2-\sqrt{3}) }{2}$

$\sf\implies 2 -\sqrt{3}$ Now comparing with $\sf a + b\sqrt3$

→ we get a = 2 and b = -1

b) Solution

→We Have

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

→Using Rationalization Method

$\implies\sf \dfrac{3+\sqrt{7} }{3-\sqrt{7}} \times \dfrac{3+\sqrt{7} }{3+\sqrt{7}}$

→using this identities

$\sf\implies ( a +b )^2 = a^2+b^2 + 2ab$

$\sf\implies( a - b)(a + b) = a^2- b^2$

→We get

$\implies\sf \dfrac{(3+\sqrt{7})^2 }{(3-\sqrt{7})(3+\sqrt{7})}$

$\sf\implies\dfrac{(3)^2+(\sqrt{7})^2+6\sqrt7}{(3)^2-(\sqrt{7})^2}$

$\sf\implies\dfrac{9+7+6\sqrt{7}}{9-7} =\dfrac{16+6\sqrt{7}}{2} =\dfrac{2(8+3\sqrt{7})}{2}$

$\sf\implies 8+3\sqrt{7}$ Now Comparing with $\sf a+b\sqrt{7}$

→ we get a = 8 and b = 3