The product of real of the equation $|x|^{6 / 5}-26|x|^{3 / 5}-27=0$

Question

Aicharya · Accepted Answer

Given equation is $|x|^{6 / 5}-26|x|^{3 / 5}-27=0$ Put $|x|^{3 / 5}=t$ $\begin{aligned} & 	herefore \quad t^2-26 t-27=0 \ & \Rightarrow \quad t^2-27 t+t-27=0 \ & \Rightarrow \quad t(t-27)+1(t-27)=0 \ & \Rightarrow \quad(t+1)(t-27)=0 \ & \Rightarrow \quad t=27 	ext { or }-1 \ & \Rightarrow \quad|x|^{3 / 5}=27 \ & \left(\because|x|^{3 / 5} 	ext { can not be negative) }ight. \ & \Rightarrow \quad|x|^3=\left(3^3ight)^5 \ & \Rightarrow \quad|x|=3^5 \ & \Rightarrow \quad x=3^5 	ext { or }-3^5 \ & \end{aligned}$ $	herefore$ Product of $x=3^5 	imes(-3)^5=-3^{10}$

The product of real of the equation $|x|^{6 / 5}-26|x|^{3 / 5}-27=0$

Solution