Determine all real numbers such that $\displaystyle (x^2-3x)^2=4-3(3x-x^2)$
let $\displaystyle u = 3x-x^2$
$\displaystyle u^2 = 4 - 3u$
$\displaystyle u^2 + 3u - 4 = 0$
$\displaystyle (u + 4)(u - 1) = 0$
first solution ...
$\displaystyle u = -4$
$\displaystyle 3x - x^2 = -4$
$\displaystyle 0 = x^2 - 3x - 4$
$\displaystyle 0 = (x - 4)(x + 1) $
$\displaystyle x = 4$ , $\displaystyle x = -1$
second solution ...
$\displaystyle u = 1$
$\displaystyle 3x - x^2 = 1$
$\displaystyle 0 = x^2 - 3x + 1$
$\displaystyle x = \frac{3 \pm \sqrt{5}}{2}$
Hello, I-Think!
That substitution is clever, but it can be solved head-on . . .
The simplified equation is: .$\displaystyle x^4 - 6x^3 + 6x^2 + 9x - 4 \;=\;0$Solve for $\displaystyle x\!:\;\; (x^2 -3x)^2 \:=\:-3(3x-x^2)$
Factor: .$\displaystyle (x+1)(x-4)(x^2-3x+1) \;=\;0$
Therefore: .$\displaystyle x \;=\;-1,\:4,\:\tfrac{3\pm\sqrt{5}}{2}$
Thanks for replying and managing to read the undecipherable text.
Here's the problem I'm having with the question.
My solution:
$\displaystyle (x^2-3x)^2=4-3(3x-x^2)$
$\displaystyle (x^2-3x)^2-4=-3(3x-x^2)$
$\displaystyle (x^2-3x+2)(x^2-3x-2)=-3(3x-x^2)$
Case 1: $\displaystyle x^2-3x+2=-3, x^2-3x-2=(3x-x^2)
$
$\displaystyle x^2-3x+2=-3\rightarrow{x^2-3x+5=0}$
$\displaystyle x=$ a complex number
$\displaystyle x^2-3x-2=3x-x^2$
$\displaystyle x^2-3x-1=0$
$\displaystyle x=\frac{3\pm{\sqrt{13}}}{2}$
Case 2: $\displaystyle x^2-3x+2=3x-x^2, x^2-3x-2=-3$
$\displaystyle x^2-3x+2=3x-x^2$
$\displaystyle x^2-3x+1=0$
$\displaystyle x=\frac{3\pm{\sqrt{5}}}{2}$
$\displaystyle x^2-3x-2=-3$
$\displaystyle x^2-3x+1=0$
Same as above
Solution Set:$\displaystyle x=\frac{3\pm{\sqrt{13}}}{2}$,$\displaystyle \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