Question

a = √3-√2/√3+√2 and b = √3+√2/√3-√2 , Find the value of a²+b² - 5ab.​

Answer 1

Explanation :

Given :

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

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

$\\ \bf a=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Rationalising the denominator,

$\implies\sf a=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Using identity (a + b)(a - b) = (a² - b²),

$\\ \implies\sf a=\dfrac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$

Using identity (a - b)² = a² - 2ab + b²,

$\\ \implies\sf a=\dfrac{(\sqrt{3})^2-2(\sqrt{3})(\sqrt{2})+(\sqrt{2})^2}{3-2}$

$\\ \implies\sf a=\dfrac{3-2\sqrt{6}+2}{1}$

$\\ \implies\sf a=3+2-2\sqrt{6}$

$\\ \therefore\boxed{\pmb{\sf a=5-2\sqrt{6}.}}$

$\\ \\ \bf b=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Rationalising the denominator,

$\implies\sf b=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Using identity (a + b)(a - b) = (a² - b²),

$\\ \implies\sf b=\dfrac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$

Using identity (a + b)² = a² + 2ab + b²,

$\\ \implies\sf b=\dfrac{(\sqrt{3})^2+2(\sqrt{3})(\sqrt{2})+(\sqrt{2})^2}{3-2}$

$\\ \implies\sf b=\dfrac{3+2\sqrt{6}+2}{1}$

$\\ \implies\sf b=3+2+2\sqrt{6}$

$\\ \therefore\boxed{\pmb{\sf b=5+2\sqrt{6}.}}$

Now,

$\bf a=5-2\sqrt{6}$

$\sf a^2=(5-2\sqrt{6})^2$

Using identity (a - b)² = a² - 2ab + b²,

$\sf a^2=(5)^2-(2)(5)(2\sqrt{6})+(2\sqrt{6})^2$

$\therefore\boxed{\pmb{\sf a^2=25-20\sqrt{6}+24.}}$

$\\ \\ \bf b=5+2\sqrt{6}$

$\sf b^2=(5+2\sqrt{6})^2$

Using identity (a + b)² = a² + 2ab + b²,

$\sf a^2=(5)^2+(2)(5)(2\sqrt{6})+(2\sqrt{6})^2$

$\therefore\boxed{\pmb{\sf b^2=25+20\sqrt{6}+24.}}$

Now,

• a = 5 - 2√6
• b = 5 + 2√6
• a² = 25 - 20√6 + 24
• b² = 25 + 20√6 + 24

$\\ =\sf a^2+b^2-5ab$

$\\ =\sf 25-20\sqrt{6}+24+25+20\sqrt{6}+24-5[(5-2\sqrt{6})(5+2\sqrt{6})]$

Using identity (a + b)(a - b) = (a² - b²),

$\\ =\sf 25-20\sqrt{6}+24+25+20\sqrt{6}+24-5[(5)^2-(2\sqrt{6})^2]$

Cancelling 20√6,

$\\ =\sf 25\ \cancel{-20\sqrt{6}}\ +24+25\ \cancel{+20\sqrt{6}}\ +24-5[25-24]$

$\\ =\sf 25+24+25+24-5[1]$

$\\ =\sf 50+48-5$

$\\ =\sf 98-5$

$\\ =\sf 93$

$\\ \\ \therefore\boxed{\pmb{\bf a^2+b^2-5ab=93.}}$