Question

please answer this question​

Answer 1

Answer:

$a = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}+\sqrt{2}}\\\\a = \frac{(\srtqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\\\a=\frac{(\sqrt{3}-\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\\\a = \frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}-2(\sqrt{3})(\sqrt{2})}{3-2}\\\\a=3+2-2\sqrt{6}\\\\a = 5-2\sqrt{6}$

$b= \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3}-\sqrt{2}}\\\\b = \frac{(\srtqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\\\b=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\\\b = \frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2(\sqrt{3})(\sqrt{2})}{3-2}\\\\b=3+2+2\sqrt{6}\\\\b = 5+2\sqrt{6}$

Thus

$(a +b)^{2} = [(5-2\sqrt{6})+(5+\sqrt{6})]^{2}\\\\(a+b)^{2}=[5-2\sqrt{6}+5+2\sqrt{6}]^{2}\\\\(a+b)^{2}=(10)^{2}\\\\(a+b)^{2}=100$

Answer 2

Answer :

(a + b)² = 100

Solution :

• Given : a = (√3 - √2)/(√3 + √2)

b = (√3 + √2)/(√3 - √2)

• To find : (a + b)² = ?

We have ;

a = (√3 - √2)/(√3 + √2)

Rationalising the denominator , we have ;

=> a = (√3-√2)(√3-√2) / (√3+√2)(√3-√2)

=> a = (√3 - √2)²/ [ (√3)² - (√2)² ]

=> a = (√3 - √2)² / (3 - 2)

=> a = (√3 - √2)² / 1

=> a = (√3 - √2)²

=> a = (√3)² - 2×√3×√2 + (√2)²

=> a = 3 - 2√6 + 2

=> a = 5 - 2√6

Also ,

b = (√3 + √2)/(√3 - √2)

Rationalising the denominator , we have ;

=> b = (√3+√2)(√3+√2) / (√3-√2)(√3+√2)

=> b = (√3 + √2)²/ [ (√3)² - (√2)² ]

=> b = (√3 + √2)² / (3 - 2)

=> b = (√3 + √2)² / 1

=> b = (√3 + √2)²

=> b = (√3)² + 2×√3×√2 + (√2)²

=> b = 3 + 2√6 + 2

=> b = 5 + 2√6

Now ,

=> a + b = 5 - 2√6 + 5 + 2√6

=> a + b = 10

Now ,

Squaring both the sides , we get ;

=> (a + b)² = 10²

=> (a + b)² = 100

Hence , (a + b)² = 100