Question

if a^2+b^2+(1/a^2)+(1/b^2) = 4 then then root of a^2 + b^2 is?btw i know this is on a exam in india that i cant talk about ;) but i have submitted already and this is my honest doubt after submitting.

Answer 1

Given,

$\longrightarrow a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}=4$

Or,

$\longrightarrow a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}=4$

Subtracting 4,

$\longrightarrow a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}-4=0$

$\longrightarrow a^2+\dfrac{1}{a^2}-2+b^2+\dfrac{1}{b^2}-2=0$

$\longrightarrow \left(a^2+\dfrac{1}{a^2}-2\cdot a\cdot\dfrac{1}{a}\right)+\left(b^2+\dfrac{1}{b^2}-2\cdot b\cdot\dfrac{1}{b}\right)=0$

$\longrightarrow \left(a-\dfrac{1}{a}\right)^2+\left(b-\dfrac{1}{b}\right)^2=0$

We know the square of every real number is always non - negative.

So if we add two squares then the sum will always be non - negative, i.e., either 0 or any positive real number.

The sum of squares of two numbers will be equal to 0 if and only if the individual numbers are equal to 0.

$\longrightarrow x^2+y^2=0\quad\iff\quad x=y=0,\ \ x,\,y\in\mathbb{R}$

Hence,

$\longrightarrow a-\dfrac{1}{a}=b-\dfrac{1}{b}=0$

$\longrightarrow \dfrac{a^2-1}{a}=\dfrac{b^2-1}{b}=0$

$\longrightarrow a^2-1=b^2-1=0$

$\longrightarrow a^2=b^2=1$

Hence,

$\longrightarrow \sqrt{a^2+b^2}=\sqrt{1+1}$

$\longrightarrow\underline{\underline{\sqrt{a^2+b^2}=\sqrt{2}}}$

Hence √2 is the answer.