Dhaka 2012 Secondary and Higher secondary

[Neblina]

Find the set of real values of c for which the equation $x^4 = (2x - c)^{2}$ has four
distinct real roots.

[Neblina]

does anyone have a solution??

[*Mahi*]

$x^4 = (2x-c)^2$ implies that $x^2 = 2x-c$ or $x^2=-2x+c$, or $x^2 -2x+c = 0$ or $x^2+2x-c = 0$, so if both of the equations have two distinct real roots, both of their discriminant should be positive and different, which means $4-4c$ and $4+4c$ should be different and positive. Which gives us the range $-1 < c < 1$, and thus the set is $c \in \left ( -1, 1 \right)$

[SANZEED]

$x^4 = (2x-c)^2$ implies that $x^2 = 2x-c$ or $x^2=-2x+c$, or $x^2 -2x+c = 0$ or $x^2+2x-c = 0$, so if both of the equations have two distinct real roots, both of their discriminant should be positive and different, which means $4-4c$ and $4+4c$ should be different and positive. Which gives us the range $-1 < c < 1$, and thus the set is $c \in \left ( -1, 1 \right)$

Mahi vai, $c=0$ gives three distinct real roots namely $0,2,-2$,not four,while the question wants $4$ distinct real roots. Shouldn't $0$ be excluded from the set?

[Phlembac Adib Hasan]

I think SANZEED is right. The equation has four roots $1\pm \sqrt {1-c},-1\pm \sqrt {1+c}$. Mahi vai, you checked $1\pm \sqrt {1-c}$ and $-1\pm \sqrt {1+c}$ pairwisely, but not all the roots.
$1+\sqrt {1-c}>1-\sqrt {1-c}\ge 0\ge -1+\sqrt {1+c}>-1-\sqrt {1+c}$.

So there could be only one case when $1-\sqrt {1-c}=0= -1+\sqrt {1+c}$ possible iff $c=0$. This is what SANZEED mentioned. So $c\in (-1,1)\backslash 0$.

[*Mahi*]

I used the line " $4−4c$ and $4+4c$ should be different and positive.", and then forgot to mention $c= 0$ makes both same my bad.