Question

The value(s) of $x$ satisfying the equation $x^9+\frac{9}{8}{x}^6+\frac{27}{64}{x}^3-x+\frac{219}{512}=0$ is/are

• A. $\frac{-1+\sqrt{13}}{4}$
• B. $-\frac{1}{2}$
• C. $\frac{-1-\sqrt{13}}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{-1+\sqrt{13}}{4}$
C $\frac{-1-\sqrt{13}}{4}$
D $\frac{1}{2}$

$x^9+\frac{9}{8}{x}^6+\frac{27}{64}{x}^3-x+\frac{219}{512}=0$
$\Rightarrow 8^3{x}^9+9\cdot 8^2{x}^6+27\cdot 8x^3-8^{3}x+219=0$
$\Rightarrow [8x^3+3{]}^3=512x-192$
$\Rightarrow x^3+\frac{3}{8}={(x-\frac{3}{8})}^{1/3}$
Now, let $f\left(x\right)=x^3+\frac{3}{8}$ where $f:R\to R$
Then, $f^{-1}\left(x\right)={(x-\frac{3}{8})}^{1/3}$
$\Rightarrow f\left(x\right)=f^{-1}\left(x\right)$
We know that $f\left(x\right)$ and $f^{-1}\left(x\right)$ is symmetrical about $y=x$ line.
$\therefore f\left(x\right)=f^{-1}\left(x\right)=x$
$\Rightarrow x^3+\frac{3}{8}=x\Rightarrow (2x-1)(4x^2+2x-3)=0$
$\Rightarrow x=\frac{1}{2},\frac{-1+\sqrt{13}}{4},\frac{-1-\sqrt{13}}{4}$