Question

please evaluate this by rationalising the denominator ​

Answer 1

Step-by-step explanation:

answer as shown above.......

Answer 2

Answer:

Hi mate

Here is ur answer ✏

$\huge\mathfrak{\underline{\underline{\red{Answer}}}}$

$\sf \frac{ \sqrt{2 } + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2 } + 2 \sqrt{3} } = x - y \sqrt{6}$

$\sf \frac{3 \times 2 + 2 \times 3}{9 \times 2 - 4 \times 3} = x - y \sqrt{6}$

$\sf \frac{12}{6} = x - y \sqrt{6}$

$\sf 2 = x - y \sqrt{6}$

$\sf \: x - \sqrt{6} y - 2 = 0$

$\sf \: x - \sqrt{6} y - 2 = 0$

$\Rightarrow\sf \: a = 1$

$\Rightarrow\sf \: b = - \sqrt{6}$

$\Rightarrow\sf \: c = - 2$

$\sf\frac{ -b + \sqrt{ {b}^{2} - 4ac } }{2a}$

$\sf = \frac{ \sqrt{6} + \sqrt{ - ({6})^{2} } - 4 \times 1 \times ( - 2)} {2 \times 1}$

$\sf = \frac{ \sqrt{6} + \sqrt{6 + 8} }{2}$

$\frac{ \sqrt{6 } + \sqrt{14} }{2}$

$\sf \frac{ \sqrt{6} + \sqrt{14} }{2} \: \: and \: \frac{ \sqrt{6} - \sqrt{14} }{2}$

$\Rightarrow\sf \: x = \frac{ \sqrt{6} + \sqrt{14} }{2}$

$\Rightarrow\sf \: y = \frac{ \sqrt{6} - \sqrt{14} }{2}$