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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestx=(a+b+c)^2

Log in to reply

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.

Log in to reply

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.....

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

Here are some observations:

Log in to reply

Log in to reply

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.

Log in to reply

## Calvin Lin plz help me.

Log in to reply

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

Log in to reply

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.

Log in to reply