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

Correct answer: A. 14 aligned & (x^2-9 x+11)^2-(x-4)(x-5)=3 & (x^2-9 x+11)^2-(x^2-9 x+20)=3 aligned Let x^2-9 x+11=t aligned & t^2-(t+9)=3 & t^2-t-12=0 & t^2-4 t+3 t-12=0 & t(t-4)+3(t-4)=0 & t=4 or -3 & x^2-9 x+11=4 & x^2-9 x+7=0 aligned Here, we will get irrational roots aligned & x^2-9 x+11=-3 & x^2-9 x+14=0 & x^2-7 x-2 x+14=0 & x=7,2 & Product of all rational roots =14 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Quadratic Equation

The product of all the rational roots of the equation (x^2-9...

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

Options

A $14$
B $21$
C $28$
D $7$

Correct answer

A. $14$

Step-by-step 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{\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\text{ of } all rational roots }=14 \end{aligned}$$

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