Question

if 7+2√2/2+3√2 =a+ b √2​

Answer 1

$\mathbb\green{GIVEN}$

• 7+2√2/2+3√2

$\mathbb\red{METHOD}$

• Use the method rationalising denominator of the given surd

$\mathbb\green{FORMULAE}$

• [ (a+b) (a-b)=a²-b² ]

• Rationalising denominator is a form of bringing the denominator root from the bottom of the fraction to top with + or - according to the root

• 7+2√2/2+3√2=a+b√2
• 7+2√2(2-3√2)/2+3√2=a+b√2
• 14-21√2+4√2-12/(2)²+(3√2)²=a+b√2
• 2-17√2/4+9×2=a+b√2
• 2-17√2/22=a+b√2

$so \: here \: we \: got \: a \: denominator \: as \: 22 \: which \: is \: \\ rational \: and \: \\ \\ \\ a + b \sqrt{2} \: is \: - 15 \sqrt{2}$

So here A is 2 and B is -17

Answer 2

Answer

a = -1/7 , b = 17/14

Explanation

$\rm \dfrac{7+2\sqrt{2}}{2+3\sqrt{2}}=a+b\sqrt{2}\\\\\implies \rm \dfrac{7+2\sqrt{2}}{2+3\sqrt{2}}\times \dfrac{2-3\sqrt{2}}{2-3\sqrt{2}}=a+b\sqrt{2}\\\\\implies \rm \dfrac{14-21\sqrt{2}+4\sqrt{2}-12}{(2)^2-(3\sqrt{2})^2}=a+b\sqrt{2}\\\\\implies \rm \dfrac{2-17\sqrt{2}}{4-18}=a+b\sqrt{2}\\\\\implies \rm \dfrac{-(17\sqrt{2}-2)}{-14}=a+b\sqrt{2}\\\\\implies \rm \dfrac{17\sqrt{2}-2}{14}=a+b\sqrt{2}\\\\\implies \rm \bigg(\dfrac{-2}{14}\bigg)+\bigg( \dfrac{17}{14}\bigg)\sqrt{2}=a+b\sqrt{2}$

$\implies \rm a=-\dfrac{1}{7},b=\dfrac{17}{14}$