# Math Help - solye 2

1. ## solye 2

2. Originally Posted by perash
Well an obvious answer is x=1, and the other obvious root is x=3.

These are obvious because for these: $\sqrt{-3+4x}=x$

From a sketch of the graphs of both sides of the equation it seems that these
are th only real roots.

RonL

3. Originally Posted by perash
$x = \sqrt{-3 + 4\sqrt{-3 + 4\sqrt{-3 + 4x}}}$

So
$x^2 = -3 + 4\sqrt{-3 + 4\sqrt{-3 + 4x}}
$

$x^2 + 3 = 4\sqrt{-3 + 4\sqrt{-3 + 4x}}
$

$(x^2 + 3)^2 = 16(-3 + 4\sqrt{-3 + 4x})
$

$x^4 + 6x^2 + 9 = -48 + 64\sqrt{-3 + 4x}
$

$x^4 + 6x^2 + 57 = 64\sqrt{-3 + 4x}
$

$(x^4 + 6x^2 + 57)^2 = 64^2(-3 + 4x)
$

$x^8 + 12x^6 + 150x^4 + 684x^2 + 3249 = -12288 + 16384x$

$x^8 + 12x^6 + 150x^4 + 684x^2 - 16384x + 12288 = 0$

(Yeach!)

Let's use the rational root theorem on this monster:
The prime factorization of 12288 is $2^{12} \cdot 3$. So possible rational roots are:
(plus or minus of all these) 1, 2, 4, 6, 8, 12, ...

We have to try all of these.

But since CaptainBlack was nice enough to solve the problem in a different way we can simply verify that x = 1 and x = 3 are the only rational solutions. (By graphing we may verify these are the only two real solutions as well.)

-Dan