Q31 / 75 · JEE Main 2025 (24 Jan Shift 1) · Mathematics

The product of all the rational roots of the equation $\left(x^2-9 x+11\right)^2-(x-4)(x-5)=3$, is equal to

  • Correct answer
    A
    $14$
  • B
    $21$
  • C
    $28$
  • D
    $7$

Solution

$\begin{aligned} & \left(x^2-9 x+11\right)^2-(x-4)(x-5)=3 \\ & \left(x^2-9 x+11\right)^2-\left(x^2-9 x+20\right)=3 \end{aligned}$ Let $x^2-9 x+11=t$ $\begin{aligned} & t^2-(t+9)=3 \\ & \Rightarrow t^2-t-12=0 \\ & \Rightarrow t^2-4 t+3 t-12=0 \\ & \Rightarrow t(t-4)+3(t-4)=0 \\ & \Rightarrow t=4 \text { or }-3 \\ & x^2-9 x+11=4 \\ & x^2-9 x+7=0 \end{aligned}$ Here, we will get irrational roots $\begin{aligned} & x^2-9 x+11=-3 \\ & x^2-9 x+14=0 \\ & x^2-7 x-2 x+14=0 \\ & \Rightarrow x=7,2 \\ & \Rightarrow \text { Product of all rational roots }=14 \end{aligned}$

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

Relevant Questions