Question

if a2 +b2+1/a2+1/b2=4 find root a2+b2

Answer 1

✪ᴀɴsᴡᴇʀ✪

$\red{\boxed{\sqrt{a^2+b^2} = \sqrt{2}} }$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Here

$\:\:\:\:\:\:\:\:\:\:a^2 +b^2+\frac{1}{a^2} +\frac{1}{b^2} =4$

$\implies [a^2 + \frac{1}{a^2}-2]$

$\:\:\:\:\:\:\:+[b^2 +\frac{1}{b^2} -2]=0$

$\implies [a^2 + \frac{1}{a^2}-2a.\frac{1}{a}]$

$\:\:\:\:\:\:\:+[b^2 +\frac{1}{b^2} -2b.\frac{1}{b}]=0$

$\implies [a- \frac{1}{a}]^2 +[b-\frac{1}{b}]^2 = 0$

Since

$[a- \frac{1}{a}]^2 ≥ 0$ and

$[b- \frac{1}{b}]^2 ≥ 0$

The above equation is possible only if both terms are zero.

$\implies [a- \frac{1}{a}]^2 = 0, [b-\frac{1}{b}]^2 = 0$

$\implies a- \frac{1}{a} = 0, b-\frac{1}{b} = 0$

$\implies a = \frac{1}{a}, b = \frac{1}{b}$

$\implies a^2 = 1, b^2 = 1$

Therefore

$\:\:\:\: \sqrt{a^2+b^2}$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

Answer 2

Answer:

A2+b2+1/A2+1/b2=4

to a2+b2 ka man hoga