$|x^2-3x+2 | = mx$ has $x_1, x_2, x_3, x_4 $ $s(m) = \frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} + \frac{1}{{x_3}^2 }+ \frac{1}{{x_4}^2}$ express $s(m)$

Question

$|x^2-3x+2 | = mx$ has $x_1, x_2, x_3, x_4 $ 4 distinct solutions $s(m) = \frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} + \frac{1}{{x_3}^2 }+ \frac{1}{{x_4}^2}$ Express $s(m)$ in terms of $m$

$0 <m < 3-2\sqrt{2}$

$(x^2-3x+2 ) = \pm mx$ I get $x^2-(3+m)x+2 = 0$ and $x^2-(3-m)x+2 = 0$

i thought that i had to find the first $\frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} =\frac{({{x_1}+{x_2}})^2 - 2x_1x_2}{({x_1x_2})^2} $ from $x^2-(3+m)x+2 = 0$ and then find $ \frac{1}{{x_3}^2} +\frac{1}{{x_4}^2}$ from $x^2-(3-m)x+2 = 0$

Answer 1

$$(x^2-3x+2)^2=m^2x^2$$

$$\implies x^4-6x^3+x^2(9+4-m^2)-12x+4=0$$

$$\iff x^4+(13-m^2)x^2+4=12x+6x^3$$

Replace $x^2=\dfrac1y\implies y=\pm\dfrac1{\sqrt x}$

$$\dfrac{1+(13-m^2)y+4y^2}{y^2}=\pm\dfrac{12y+6}{y^{3/2}}$$

$$(1+(13-m^2)y+4y^2)^2= y(12y+6)^2$$

$$\implies 16y^4+y^3(8(13-m^2)-144)+\cdots+1=0$$

Using Vieta's formula $$\sum_{r=1}^4\dfrac1{x_r^2}=\sum_{r=1}^4y_r=\dfrac{8m^2+40}{16}$$