1. ## Question on method

Determine all real numbers such that $(x^2-3x)^2=4-3(3x-x^2)$

3. Problem fixed.

4. Originally Posted by I-Think
Determine all real numbers such that $(x^2-3x)^2=4-3(3x-x^2)$

let $u = 3x-x^2$

$u^2 = 4 - 3u$

$u^2 + 3u - 4 = 0$

$(u + 4)(u - 1) = 0$

first solution ...

$u = -4$

$3x - x^2 = -4$

$0 = x^2 - 3x - 4$

$0 = (x - 4)(x + 1)$

$x = 4$ , $x = -1$

second solution ...

$u = 1$

$3x - x^2 = 1$

$0 = x^2 - 3x + 1$

$x = \frac{3 \pm \sqrt{5}}{2}$

5. Hello, I-Think!

That substitution is clever, but it can be solved head-on . . .

Solve for $x\!:\;\; (x^2 -3x)^2 \:=\:-3(3x-x^2)$
The simplified equation is: . $x^4 - 6x^3 + 6x^2 + 9x - 4 \;=\;0$

Factor: . $(x+1)(x-4)(x^2-3x+1) \;=\;0$

Therefore: . $x \;=\;-1,\:4,\:\tfrac{3\pm\sqrt{5}}{2}$

Here's the problem I'm having with the question.

My solution:
$(x^2-3x)^2=4-3(3x-x^2)$
$(x^2-3x)^2-4=-3(3x-x^2)$
$(x^2-3x+2)(x^2-3x-2)=-3(3x-x^2)$

Case 1: $x^2-3x+2=-3, x^2-3x-2=(3x-x^2)
$

$x^2-3x+2=-3\rightarrow{x^2-3x+5=0}$
$x=$ a complex number
$x^2-3x-2=3x-x^2$
$x^2-3x-1=0$
$x=\frac{3\pm{\sqrt{13}}}{2}$

Case 2: $x^2-3x+2=3x-x^2, x^2-3x-2=-3$
$x^2-3x+2=3x-x^2$
$x^2-3x+1=0$
$x=\frac{3\pm{\sqrt{5}}}{2}$

$x^2-3x-2=-3$
$x^2-3x+1=0$
Same as above

Solution Set: $x=\frac{3\pm{\sqrt{13}}}{2}$, $\frac{3\pm{\sqrt{5}}}{2}
$

Now my problem is, the official solution to this question gives the same answer as skeeter and soroban, but my method gives two different answers.
But the thing is, they seem to be equally valid, I verified it with a calculator, and I'm wondering if this is possible.
Can a quartic curve intersect with a quadratic curve 6 times?
This is the question I would like answered.

Thanks

7. the two solutions , $\frac{3 \pm \sqrt{13}}{2}$ , do not work in the original equation ...

what's that tell you?

8. My calculator has betrayed me...
My apologies for wasting the forum's time due to a calculator error.