Q89 / 160 · AP EAMCET 2012 · Mathematics

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

  • Correct answer
    A
    $-3^{10}$
  • B
    $-3^{12}$
  • C
    $-3^{12 / 5}$
  • D
    $-3^{21 / 5}$

Solution

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

Hence, the correct answer is option: 'A'.

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