Verified Solution • 2025–26 Session moderate

QJEEM25Ma7f03: JEE Main 2025 Mathematics - Class 11 JEEM25Ma7f03 Solution: The product of all the rational roots of the equation , is…

The product of all the rational roots of the equation (x2−9x+11)2−(x−4)(x−5)=3\left(x^2-9 x+11\right)^2-(x-4)(x-5)=3 , is equal to

Multiple Choice Options

  • A 7
  • B 21
  • C 28
  • D 14

Step-by-Step Verified Solution

amp;(x2−9x+11)2−(x2−9x+20)=3amp; Let amp;amp;⇒x2−9x=tamp;⇒t2+22t+121−t−20−3=0amp;⇒t2+21t+98=0amp;⇒(t+14)(t+7)=0amp;⇒t=−7,−14\begin{aligned}&\left(x^2-9 x+11\right)^2-\left(x^2-9 x+20\right)=3\\&\text { Let }\\&\begin{aligned}& \Rightarrow \mathrm{x}^2-9 \mathrm{x}=\mathrm{t} \\& \Rightarrow \mathrm{t}^2+22 \mathrm{t}+121-\mathrm{t}-20-3=0 \\& \Rightarrow \mathrm{t}^2+21 \mathrm{t}+98=0 \\& \Rightarrow(\mathrm{t}+14)(\mathrm{t}+7)=0 \\& \Rightarrow \mathrm{t}=-7,-14\end{aligned}\end{aligned}

amp; So, amp;amp;x2−9x=−7,−14amp;x2−9x+7=0 or x2−9x+14=0\begin{aligned}&\text { So, }\\&\begin{aligned}& x^2-9 x=-7,-14 \\& x^2-9 x+7=0 \quad \text { or } \quad x^2-9 x+14=0\end{aligned}\end{aligned}

amp;x=9±81−4(7)2×1amp;x=9±81−4(14)2=9±532amp;=9±52amp; Product of all rational roots =7×2=14\begin{aligned}&\begin{array}{ll}x=\frac{9 \pm \sqrt{81-4(7)}}{2 \times 1} & x=\frac{9 \pm \sqrt{81-4(14)}}{2} \\=\frac{9 \pm \sqrt{53}}{2} & =\frac{9 \pm 5}{2}\end{array}\\&\text { Product of all rational roots }=7 \times 2=14\end{aligned}

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