Question

$\frac{ {x}^{2} - 4 {a}^{2} - 2 {c}^{2} }{a + b} + \frac{ {x}^{2} - 4 {b}^{2} - 2 {a}^{2} }{b + c} + \frac{ {x}^{2} - 4 {c}^{2} - 2 {b}^{2} }{c + a} = 0 \\ what \: is \: the \: value \: of \: x$
the answer is. +- root 2(a^+b^+c^)


please explain explain............
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Answer 1

Answer:

I am adding my calculations hope that helps

Take lcm

Let

$(a + b)(a + c) = p \\ (b + c)(b + a) = q \\ (c + a)(c + b) = r$

$4 {a}^{2} (a + c)(b + c) + 2 {a}^{2}( b + a)(c + a) \\ = 2 {a}^{2} (a + c)(a + 3b + 2c) \\ = 2 {a}^{2} (p + q + r)$

Similarly other terms

and your fraction becomes

$\frac{ {x}^{2}(p + q + r) - 2 {a}^{2} (p + q + r) - 2 {b}^{2} (q + p + r) - 2 {c}^{2} (r + q + p) }{ \sqrt{pqr} } = 0 \\ \\ (p + q + r)( {x}^{2} - 2 {a}^{2} - 2{b}^{2} - 2 {c}^{2} ) = 0 \\ x = \pm \sqrt{2( {a}^{2} + {b}^{2} + {c}^{2} )}$

Answer 2

Answer:

hii nandita i am extremely sorry i didn't get time to tell u the answer okay...

Step-by-step explanation:

it was really very easy.. haha...

see u'll laugh when u'll see the answer..

i had solved it at that instant but father took the mobile away ..so u know...

okay so here we are done with it...

just follow the attachment and everything will be as clear as water i guess..:)

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