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

Answer 1

$\displaystyle\huge\purple{\underline{\underline{Question}}}$

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 ,

$\implies\:a^2\:+\:ab\:+\:b^2$

$\implies\:(3\:+\:2\sqrt{2})\:+\:1\:+\:(3\:-2\sqrt{2})$

$\implies\:9\:+\:12\sqrt{2}\:+\:8\:+\:1\:+\:9\:-\:12\sqrt{2}\:+\:8$

$\implies\:9\:+\:8\:+\:1\:+\:9\:+\:8$

${\boxed{\red{35\:is\:the\: solution}}}$