Question

Solve the equations:
$32x^3-48x^2+22x-3=0$, the roots being in arithmetical progression.

Answer 1

$32x^3-48x^2+22x-3=0$

Let the roots be $a-d,a,a+d$

$S_1=a-d+a+a+d=-\frac{-48}{32}3a=\frac{3}{2}a=\frac{1}{2}{S}_3=a(a-d)(a+d)=-\frac{-3}{32}\frac{1}{2}(\frac{1}{4}-d^2)=\frac{3}{32}\frac{1}{4}-d^2=\frac{3}{16}\frac{1}{4}-\frac{3}{16}=d^2{d}^2=\frac{4-3}{16}\Rightarrow d=\pm \frac{1}{4}$

So the roots of the equation are $\frac{1}{4},\frac{1}{2},\frac{3}{4}$