Checking for real roots in biquadratic polynomial.

Question

What are the value of $k \in \mathbb{R}$ for which the equation $(x^2-2x)^2-3(x^2-2x)+(k+2)=0$ has two real solutions.

Answer 1

As $y=x^2-2x=(x-1)^2-1\ge0-1,$

For two real solutions, we need $y_1\ge-1,y_2<-1$ for the quadratic equation

$$y^2-3y+k+2=0$$

Again $y_2=3-y_1\le3+1$ which is already true

$k+2=y_1y_2=y_1(3-y_1)=\dfrac94-(y_1-\dfrac32)^2$

Finally $y_1-\dfrac32\ge-1-\dfrac32$

Answer 2

HINT.- There is an axis of vertical symmetry because we notice that $$f (1 + x) = (x ^ 2-1) ^ 2-3 (x ^ 2-1) + k + 2 = f (1-x)$$ so the axis is given by $x = 1$. As the equation is biquadratic there are then generally two possibilities, which are no real roots or four real roots. The exceptional case of only two real roots is then given with the two minimums of the curve being equal to zero.

The derivative is $f'(x)=2(x-1)(2x^2-4x-3)$ and the minimums are taken in $x=\dfrac{2\pm\sqrt{10}}{2}$ and the asked value of $k$ is given by the equation $$\left(\left(\dfrac{2+\sqrt{10}}{2}\right)^2-2\left(\dfrac{2+\sqrt{10}}{2}\right)\right)^2-3\left(\dfrac{2+\sqrt{10}}{2}\right)^2+6\left(\dfrac{2+\sqrt{10}}{2}\right)+k+2=0$$ which gives $$-0.25+k=0$$