discriminant

• Apr 10th 2011, 03:45 AM
Punch
discriminant
prove algebraically that no portion of the curve $\displaystyle y=\frac{(a-x)^2}{2-x}$ exists $\displaystyle for 0<y<4a-8$
• Apr 10th 2011, 04:00 AM
FernandoRevilla
$\displaystyle f(x)=y\Leftrightarrow \ldots \Leftrightarrow x^2+(-2a+y)x+a^2-2y=0$

There exists $\displaystyle x$ iff

$\displaystyle \Delta=\ldots=y(y+8-4a)\geq 0$

Now, you can conclude.
• Apr 11th 2011, 02:20 AM
Punch
Quote:

Originally Posted by FernandoRevilla
$\displaystyle f(x)=y\Leftrightarrow \ldots \Leftrightarrow x^2+(-2a+y)x+a^2-2y=0$

There exists $\displaystyle x$ iff

$\displaystyle \Delta=\ldots=y(y+8-4a)\geq 0$

Now, you can conclude.

$\displaystyle b^2-4ac<0$,

$\displaystyle (-2a+y)^2-4(1)(a^2-2y)<0$
$\displaystyle y^2-4ay+8y<0$
$\displaystyle y(y-4a+8)<0$

$\displaystyle y<0$ or $\displaystyle y<4a-8$
but this does not tally with the one in the question
• Apr 11th 2011, 02:49 AM
FernandoRevilla
$\displaystyle (y>0)\;\wedge\;(y<4a-8)\Rightarrow (y>0)\;\wedge\;(y-4a+8)<0\Rightarrow \Delta<0$