Question

triangle.
8. Rationalize the denominator of the irrational number √3+√2/√3-√2
​

Answer 1

Hy

Here is ur answer

√3+√2/√3-√2*√3+√2/√3+√2

=(√3+√2)power 2/(√3-√2) (√3+√2)

=(√3)power2+(√2)power2 +2*√3*√2/(√3)power2-(√2)power2

=3+2+2√6/3-2

=5+2√6/1

=5+2√6. ans.

I hope it helps u .

Answer 2

$\underline{\underline{\textbf{\huge{Question}}}}$

Rationalise

$\mathbf{\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } }$

$\underline{\underline{\textbf{\huge{Answer}}}}$

As we Know

√a+√b is Rationalise Factor of √a-√b

Then √3+√2 is Rationalise Factor of √3-√2

Then Multiply √3+√2 with Numerator and Denominator.

$\mathbf{ \frac{ \sqrt{3} + \sqrt{2} \times \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} \times \sqrt{3} + \sqrt{2} } } \\ \\ \mathbf{ \frac{( \sqrt{3} + \sqrt{2} ) {}^{2} }{( \sqrt{3}) {}^{2} - ( \sqrt{2}) {}^{2} }}$


As we know (a+b)²=a²+b²+2ab


a²-b²=(a-b)(a+b)


(√3+√2)²=√3²+√2²+2×√3×√2


$\mathbf{ \frac{ (\sqrt{3} ) {}^{2} + ( \sqrt{2}) {}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{ \sqrt{3} \times \sqrt{3} - \sqrt{2} \times \sqrt{2} } } \\ \\ \mathbf{\frac{ \sqrt{3} \times \sqrt{3} \ + \sqrt{2} \times \sqrt{2} + 2 \times \sqrt{3 \times 2} }{ \sqrt{9} - \sqrt{4} } } \\ \\ \mathbf{ \frac{ \sqrt{9} + \sqrt{4} + 2 \sqrt{6} }{3 - 2} } \\ \\ \mathbf{\frac{3 + 2 + 2\sqrt{6} }{1} } \\ \\ \mathbf{5 + 2 \sqrt{6} }$

$\boxed{\mathbf{\huge{Answer=5+2\sqrt{6}}}}$