Question

3 + root 2 / 3 - root 2 = a + b root 2

Answer 1

Step-by-step explanation:

$\small\fbox\blue {Answer in attachment ⤴️}$

Answer 2

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:\dfrac{3+\sqrt2}{3-\sqrt2}=a+b\sqrt2$

To Find:

• Value of a and b

Solution:

$\text{We are given that,}\\\\:\longrightarrow\dfrac{3+\sqrt2}{3-\sqrt2}=a+b\sqrt2\\\\\text{Simplifying LHS,}\\\\:\implies\dfrac{3+\sqrt2}{3-\sqrt2}\\\\\text{Rationalising the denominator,}\\\\:\implies\dfrac{3+\sqrt2}{3-\sqrt2}\times\dfrac{3+\sqrt2}{3+\sqrt2}\\\\:\implies\dfrac{(3+\sqrt2)\times(3+\sqrt2)}{(3-\sqrt2)\times(3+\sqrt2)}\\\\\text{We know that,}\\\\:\hookrightarrow(x-y)(x+y)=x^2-y^2\\\\\text{So,}\\\\:\implies\dfrac{(3+\sqrt2)^2}{(3)^2-(\sqrt2)^2}\\\\\text{We know that,}\\\\:\hookrightarrow(x+y)^2=x^2+2xy+y^2\\\\:\implies\dfrac{3^2+2(3)(\sqrt2)+(\sqrt2)^2}{9-2}\\\\:\implies\dfrac{9+6\sqrt2+2}{7}\\\\:\implies\dfrac{11+6\sqrt2}{7}\\\\\text{So,}\\\\:\implies\dfrac{11+6\sqrt2}{7}=a+b\sqrt2\\\\:\implies\dfrac{11}{7}+\dfrac{6\sqrt2}{7}=a+b\sqrt2\\\\\text{Hence, on comparing like terms}\\\\\bf{:\implies a=\dfrac{11}{7}\:\:and\:\:b=\dfrac{6}{7}}$

Formula Used:

• (x-y)(x+y)=x^2 - y^2
• (x+y)^2=x^2 + 2xy + y^2