Question

root5+root3/2root5-3root3=a-broot15​

Answer 1

Solution :

$:\implies \sf{\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}} = a - b\sqrt{15}}$

From the above equation, we get that :

• $:\implies \sf{RHS = a - b\sqrt{15}}$

• $\sf{LHS = \dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}}}$

By solving the LHS of the equation, we get :

$:\implies \sf{\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}}}$

By multiplying $\sf{(2\sqrt{5} + 3\sqrt{3})}$ to both the numerator and denominator of the equation, we get :

$:\implies \sf{\dfrac{\sqrt{5} + \sqrt{3} \times (2\sqrt{5} + 3\sqrt{3})}{2\sqrt{5} - 3\sqrt{3} \times (2\sqrt{5} + 3\sqrt{3})}}$

$:\implies \sf{\dfrac{\sqrt{5}(2\sqrt{5} + 3\sqrt{3}) + \sqrt{3}(2\sqrt{5} + 3\sqrt{3})}{2\sqrt{5}(2\sqrt{5} + 3\sqrt{3}) - 3\sqrt{3}(2\sqrt{5} + 3\sqrt{3})}}$

$:\implies \sf{\dfrac{2\sqrt{5 \times 5} + 3\sqrt{3 \times 5} + 2\sqrt{3 \times 5} + 3\sqrt{3 \times 3}}{2\sqrt{5} \times 2\sqrt{5} + 2\sqrt{5} \times 3\sqrt{3} - 3\sqrt{3} \times 2\sqrt{5} - 3 \sqrt{3} \times 3\sqrt{3}}}$

$:\implies \sf{\dfrac{2 \times 5 + 3\sqrt{15} + 2\sqrt{15} + 3 \times 3}{20 + 6\sqrt{15} - 6\sqrt{15} - 27}}$

$:\implies \sf{\dfrac{10 + 3\sqrt{15} + 2\sqrt{15} + 9}{20 - 27}}$

$:\implies \sf{\dfrac{19 + 5\sqrt{15}}{-7}}$

$:\implies \sf{\dfrac{19}{-7} + \dfrac{5\sqrt{15}}{-7}}$

$\boxed{\therefore \sf{\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}} = \dfrac{19}{-7} + \dfrac{5\sqrt{15}}{-7}}}$

Now by putting the LHS and RHS together, we get :

$:\implies \sf{\dfrac{19}{-7} + \dfrac{5\sqrt{15}}{-7} = a - b\sqrt{15}}$

$:\implies \sf{\dfrac{19}{-7} = a, \dfrac{5\sqrt{15}}{-7} = -b\sqrt{15}}$

$:\implies \sf{\dfrac{19}{-7} = a, \dfrac{5\sqrt{15}}{-7 \times \sqrt{15}} = -b}$

$:\implies \sf{\dfrac{19}{-7} = a, \dfrac{5}{-7} = -b}$

$:\implies \sf{\dfrac{19}{-7} = a, \dfrac{5}{7} = b}$

$\boxed{\therefore \sf{a = \dfrac{19}{-7} \:and\: b = \dfrac{5}{7} = b}}$