Question

please answer this question ​

Answer 1

Solution:-

First, we are going to rationalise the denominator and then we can find the values of a and b

$\to \sf \dfrac{30}{5\sqrt{3}-3\sqrt{5} }=a\sqrt{3}-b\sqrt{5}$

$\to \sf \dfrac{30(5\sqrt{3} + 3\sqrt{5})}{(5\sqrt{3} - 3\sqrt{5})(5\sqrt{3} + 3\sqrt{5})} = a\sqrt{3} - b\sqrt{5}$

We know that $\sf (a+b)(a-b) = a^2 - b^2$

$\to \sf \dfrac{150\sqrt{3}+90\sqrt{5}}{(5\sqrt{3})^2 - (3\sqrt{5})^2} = a\sqrt{3} - b\sqrt{5}$

$\to \sf \dfrac{150\sqrt{3}+90\sqrt{5} }{25\times3 - 9\times5} = a\sqrt{3}-b\sqrt{5}$

$\to \sf \dfrac{150\sqrt{3}+90\sqrt{5} }{30} = a\sqrt{3}-b\sqrt{5}$

$\to \sf 150\sqrt{3}+90\sqrt{5} = 30(a\sqrt{3}-b\sqrt{5})$

Therefore,

$\to \sf 150\sqrt{3}=30a\sqrt{3} \implies 150 = 30a \implies a = 5$

$\to \sf 90\sqrt{5}=-30b\sqrt{5} \implies 90 = -30b \implies b = -3$

$\boxed{\sf a = 5, b = -3}$

Extra - Information:-

• Whenever you strick across a problem which has the format of the question like if,

$\sf a\sqrt{x} + b\sqrt{y} = c_1\sqrt{x} + c_2\sqrt{y}$

Then you have equate  the terms which have $\sf \sqrt{x}$ and the terms which have $\sf \sqrt{y}$

• While rationalising the denominator you have to multiply and divide by its conjugate
• To find the conjugate you have just change the sign between the terms of the denominator

The conjugate of a + b is a - b