Arithmetical progression and quadratic equation

Question

Determine the real number $k$ with the condition that the roots of the equation $x^ {4}-(3k+2) x^ {2} +k^ {2} =0$ make the arithmetic progression?

I dont know how to start ?

Answer 1

HINT:

Define $u=x^2$ and solve the quadratic in terms of $u$.

Then replace $u$ by $x=\sqrt u$ and you should get $4$ roots. Now use the AP rule.

Answer 2

Hint: Make the ansatz $$x_1,x_1+d,x_1+2d,x_1+3d$$ in your equation

Answer 3

Making the change: $t=x^2$ we get: $$t^2-(3k+2)t+k^ {2} =0 \Rightarrow \\ t_1=\frac{3k+2-\sqrt{5k^2-12k+4}}{2} \ \ \text{and} \ \ t_2=\frac{3k+2+\sqrt{5k^2-12k+4}}{2}.$$ Note that the equation has a real solution for: $5k^2-12k+4\ge 0$. It if does, then $t_{1,2}>0$.

Hence: $$x_1=-\sqrt{\frac{3k+2-\sqrt{5k^2-12k+4}}{2}} \ \text{and} \ x_2=\sqrt{\frac{3k+2-\sqrt{5k^2-12k+4}}{2}} \\ x_3=-\sqrt{\frac{3k+2+\sqrt{5k^2-12k+4}}{2}} \ \text{and} \ x_4=\sqrt{\frac{3k+2+\sqrt{5k^2-12k+4}}{2}}.$$ As the AP can not alternate and APs are symmetric, it is needed to check: $x_3,x_1,x_2,x_4.$

Answer 4

Let the roots be $-3d, -d, d, 3d$.

Hence $$\begin{align} x^4-(3x+2)x^2+k^2&=(x-3d)(x-d)(x+d)(x+3d)\\ &=(x^2-d^2)(x^2-3d^2)\\ &=x^4-10d^2x^2+9d^4\\ \end{align}$$ Equating coefficients of $x^2, x^0$ gives $k=\pm 3d^2$ and $10d^2=3k+2$.

Solving these gives $d^2=2, \frac 2{19}$ and $k=\pm 6, \pm \frac 6{19}$.

The solution corresponds to the case where $\color{red}{k=6}$ (i.e. $d^2=2$), in which case the equation reduces to $$ \begin{align} x^4-20k^2+36&=0\\ (x^2-2)(x^2-18)&=0 \end{align}$$ which have roots $-3\sqrt2, -\sqrt2, \sqrt2, 3\sqrt2$ which are in AP with common difference $D=2d=2\sqrt2$.