Question

3+root2 by3-root2=a+b root2 then find a,b

Answer 1

Answer:

here is your answer

Step-by-step explanation:

Solving LHS:

\small{= > \frac{3+\sqrt2}{3-\sqrt2}}=>3−23+2

Divide and multiply by \small{\frac{3+\sqrt2}{3+\sqrt2}}3+23+2

\begin{gathered}\small{= > \frac{3+\sqrt2}{3-\sqrt2}\times \frac{3+\sqrt2}{3+\sqrt2} }\\\\\small{= > \frac{(3+\sqrt2)^2}{(3+\sqrt2)(3-\sqrt2)} } \\\\\small{= > \frac{3^2+(\sqrt2)^2+2(3)(\sqrt2)}{3^2 - (\sqrt2)^2} }\\\\\small{= > \frac{9+2+6\sqrt2}{9-2}}\\\\\small{= > \frac{11+6\sqrt2}{7}} \end{gathered}=>3−23+2×3+23+2=>(3+2)(3−2)(3+2)2=>32−(2)232+(2)2+2(3)(2)=>9−29+2+62=>711+62

= > \frac{11}{7}+\small{\frac{6\sqrt2}{7}}=>711+762

Matching the values

Then \frac{6\sqrt2}{7}762 =b√2 → (6/7) = b

→ 11/7 =a