\[\frac {x-b^2}{a+c} + \frac {x-c^2}{a+b} + \frac {x-a^2}{b+c} = 4(a+b+c)\]

How can I solve this equation for \(x\)? Please show me the process and also the basic idea of solving this type of equation.

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## Comments

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TopNewestx=(a+b+c)^2

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What have you tried? Where did you get stuck?

Given that this is a linear equation in \(x\), it should be simple to shift terms around. (Of course, there may be other approaches.)

There is no guarantee that the answer you obtain will be "nice". Though, if you saw it in an olympiad, then it would likely be nice.

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My answer is not "nice". A big one. But it's an Olympiad question according to my friend ( He gave me this problem to solve). So.....

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Can you state your answer? Ideally, make it as simplified as possible.

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I agree with the denominator being \( (a+b)(b+c) + (b+c)(c+a) + (c+a)(a+b) \). It is more natural to write it as \( a^2 + b^2 + c^3 + 3ab + 3bc + 3ca \).

I'm not sure how you arrived at the numerator though. To me, it should be \( b^2(b+c)(b+a) + c^2 (c+a)(c+b) + a^2(a+b)(a+c) + 4(a+b+c)(a+b)(b+c)(c+a) \).

As always, if you lay out your steps, someone would be able to see what you did and provide feedback from there. But otherwise, we can only guess at what you're doing.

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Other than making a lucky observation about the answer, how else can we appraoch this problem?

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Here are some observations:

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It is clear that \( x = (a+b+c)^2 \) satisfies \( x = c^2 + k_c (a+b) = a^2 + k_a (b+c) = b^2 + k_b( c+a) \), but it's not clear how we can easily backtrack.

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## Calvin Lin plz help me.

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Instead of asking people for help, tell us where you're stuck because we don't know what issues you're facing.

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I didn't solve this kind of equation before. I'm trying. But couldn't find any way. So please help me to solve this.

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