1. ## Equation

solve for x the equation :
$\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+b^2}$

2. Originally Posted by beq!x
solve for x the equation :
$\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+b^2}$
Take x from both sides and square:

$\displaystyle a^2+x^2 = (\frac{5a^2}{ a^2+b^2} - x)^2 = \frac{25a^4}{(a^2+b^2)^2} - \frac{10a^2x}{a^2+b^2} + x^2$

$\displaystyle x^2$ will cancel. Rearranging gives:

$\displaystyle \frac{10a^2}{a^2+b^2}x = \frac{25a^4}{(a^2+b^2)^2}$

From there solve for x

3. Originally Posted by beq!x
$\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+b^2}$
here is a mistake it is not $\displaystyle b$ but $\displaystyle x$, sorry.
the right equation: $\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+x^2}$

i have a question for e^(i*pi) : did you forget $\displaystyle a^2$ here : $\displaystyle \frac{10a^2}{a^2+b^2}x = \frac{25a^4}{(a^2+b^2)^2}$ or am i wrong?

4. Subract x from both sides and then square both sides. I think that is how I would do it.

5. ## root questions

if $\displaystyle \sqrt{a^{2}+x^{2}}$ can we do this$\displaystyle \sqrt{a^2} + \sqrt{x^2}= a+ x$
or what should i do ???

6. Originally Posted by beq!x
if $\displaystyle \sqrt{a^{2}+x^{2}}$ can we do this$\displaystyle \sqrt{a^2} + \sqrt{x^2}= a+ x$
No, you can't!

7. Originally Posted by beq!x
if $\displaystyle \sqrt{a^{2}+x^{2}}$ can we do this$\displaystyle \sqrt{a^2} + \sqrt{x^2}= a+ x$
or what should i do ???
Does your conjecture hold when a = 1 and x = 1? (This is called a counter-example).