If a = (√2 + 1) ÷( √2 – 1) and b = (√2 - 1) ÷( √2 + 1), th

1.	If a = (√2 + 1) ÷( √2 – 1) and b = (√2 - 1) ÷( √2 + 1), then a² + ab + b² =__________. a:-34b:-35c:-30d:-33
Answer» $\displaystyle\huge\purple{\underline{\underline{Question}}}$ If a = (√2 + 1) ÷( √2 – 1) and b = (√2 - 1) ÷( √2 + 1), then a² + ab + b² =__________. a:-34 b:-35 c:-30 d :- 33 $\displaystyle\huge\red{\underline{\underline{Solution}}}$ $\sf{\underline{\green{Given}}}$ The VALUE of a = $\frac{\sqrt{2}\:+\:1}{\sqrt{2}\:-\:1}$ Rationalizing the DENOMINATOR $\implies\:\frac{(\sqrt{2}\:+\:1)\:(\sqrt{2}\:-\:1)}{(\sqrt{2}\:+\:1)\:(\sqrt{2}\:-\:1)}$ $\implies\:\frac{\sqrt{2}^2\:+\:1^2}{\sqrt{2^2}\:-\:1}$ The value of b = $\frac{\sqrt{2}\:-\:1}{\sqrt{2}\:+\:1}$ $\implies\:\frac{(\sqrt{2}\:-\:1)\:(\sqrt{2}\:+\:1)}{(\sqrt{2}\:-\:1)\:(\sqrt{2}\:+\:1)}$ $\implies\:\frac{\sqrt{2}^2\:-\:1^2}{\sqrt{2^2}\:-\:1}$ To find the value of $a^2\:+\:ab\:+\:b^2$ $\sf{\underline{\pink{Using\: algebraic\: identities}}}$ $(\:x\:+\:y\:)^2\:=\:x^2\:+\:y^2\:+2xy$ $(\:x\:-\:y\:)(\:x+\:y\:)\:=\:x^2\:-\:y^2$ $\sf{\underline{\pink{Value\:of\:a}}}$ $\leadsto\:\frac{2\:+\:1\:+\:2\sqrt{2}}{2\:-\:1}$ EQUATION 1 a $\leadsto\:3\:+\:2\sqrt{2}$ $\sf{\underline{\pink{Value\:of\:b}}}$ $\leadsto\:\frac{2\:+\:1\:-\:2\sqrt{2}}{2\:-\:1}$ Equation 2 b $\leadsto\:3\:-\:2\sqrt{2}$ Equation 3 ab $\rightarrow\:(\:3\:+\:2\sqrt{2})(\:3\:-\:2\sqrt{2})$ $\rightarrow\:3^2\:-\:(\:2\sqrt{2})$ $\rightarrow\:9\:-\:8$ $\rightarrow\:1$ Let's put these VALUES in this equation , $\implies\:a^2\:+\:ab\:+\:b^2$ $\implies\:(3\:+\:2\sqrt{2})\:+\:1\:+\:(3\:-2\sqrt{2})$ $\implies\:9\:+\:12\sqrt{2}\:+\:8\:+\:1\:+\:9\:-\:12\sqrt{2}\:+\:8$ $\implies\:9\:+\:8\:+\:1\:+\:9\:+\:8$ ${\boxed{\red{35\:is\:the\: solution}}}$