Question

4+√2/2+√2=a-√b
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Answer 1

$\textbf{Given:}$

$\dfrac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$

$\textbf{To find:}$

$\text{The values of a and b}The values of a and b$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$

$\text{To rationalize the denominator multiply both numerator}$

$\text{denominator by 2-\sqrt{2} }$

$\dfrac{4+\sqrt{2}}{2+\sqrt{2}}{\times}\dfrac{2-\sqrt{2}}{2-\sqrt{2}}=a-\sqrt{b}$

$\dfrac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=a-\sqrt{b}$

$\dfrac{8-4\sqrt{2}+2\sqrt{2}-2}{4-2}=a-\sqrt{b}$

$\dfrac{6-2\sqrt{2}}{2}=a-\sqrt{b}$

$\dfrac{2(3-\sqrt{2})}{2}=a-\sqrt{b}$

$3-\sqrt{2}=a-\sqrt{b}$

$\text{Comparing on bothsides, we get}$

$a=3\;\text{and}\;b=2a=3and b=2$

$\therefore\textbf{The value of a and b are 3 and 2}$

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