Question

if a^2 +b^2+1/a^2+1/b^2=4 find root a^2+b^2?

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

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

$= \sqrt{1 + 1}$

$= \sqrt{2}$