Question

Solve the equations:
$27x^4-195x^3+494x^2-520x+192=0$, the roots being in geometrical progression.

Answer 1

Given equation, $27x^4-195x^3+494x^2-520x+192=0$ and let the roots be $a,b,c,d$

The roots are in Geometric Progression, therefore

$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}⟹ad=bc$

We know that $abcd=\frac{192}{27}⟹ad=bc=\frac{8}{3}$

$(x-a)(x-d)=x^2-(a+d)x+ad=x^2–Ax+\frac{8}{3}$, where $A=a+d$

$(x-b)(x-c)=x^2-(b+c)x+bc=x^2–Bx+\frac{8}{3}$, where $B=b+c$

$x^4-\frac{65}{9}{x}^3+\frac{494}{27}{x}^2-\frac{520}{27}x+\frac{64}{9}=(x-a)(x-b)(x-c)(x-d)$

$=(x^2–Ax+\frac{8}{3})(x^2–Bx+\frac{8}{3})\dots .............\dots \left(1\right)$

$=x^4–(A+B)x^3+(AB+\frac{16}{3})x^2-\left(\frac{8}{3}\right)(A+B)x+\frac{64}{9}$

Comparing the coefficients of $x$ and $x^2$, we have

$A+B=\frac{520}{27}\cdot \frac{3}{8}=-\frac{65}{9}$ and $AB=-\frac{494}{27}-\frac{16}{3}=\frac{350}{27}$

Eliminating $B$ in the above equations, we have

$27A^2+195A+350=0$

$⟹(3A+10)(9A+35)=0⟹A=-\frac{35}{9}andB=-\frac{10}{3}$

Substituting value of $A$ and $B$ in (1), we have

$(x^2+\frac{35}{9}x+\frac{8}{3})(x^2+\frac{10}{3}x+\frac{8}{3})=0$

$⟹(9x^2+35x+24)(3x^2+10x+8)=0$

$⟹(x+3)(9x+8)(3x+4)(x+2)=0$

$\therefore$ roots of the given equation are: $-3,-2,-\frac{4}{3},-\frac{8}{9}$