The product of all the rational roots of the equation , is equal to [2025]

Question

The product of all the rational roots of the equation ${(x^{2} - 9 x + 11)}^{2} - (x - 4) (x - 5) = 3$ , is equal to [2025]

Smart Achievers · Accepted Answer

(4)

We have, ${(x^{2} - 9 x + 11)}^{2} - (x - 4) (x - 5) = 3$

$\Rightarrow {(x^{2} - 9 x + 11)}^{2} - (x - 9 x + 20) = 3$

Let $x^{2} - 9 x = t$ , then

${(t + 11)}^{2} - (t + 20) = 3$

$\Rightarrow t^{2} + 121 + 22 t - t - 20 - 3 = 0$

$\Rightarrow t^{2} + 21 t + 98 = 0 \Rightarrow (t + 14) (t + 7) = 0$

$\Rightarrow t = - 7, - 14$

$\Rightarrow x^{2} - 9 x + 7 = 0 and x^{2} - 9 x + 14 = 0$

$\Rightarrow x = \frac{9 \pm \sqrt{53}}{2} and x = \frac{9 \pm 5}{2}$

$\Rightarrow$ Product of all rational roots = $(\frac{9 + 5}{2}) (\frac{9 - 5}{2}) = 7 \times 2 = 14$ .

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