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(A) Given equation: $(x^2-9x+11)^2-(x^2-9x+20)=3$ Let $t = x^2-9x$. Substituting $t$ into the equation: $(t+11)^2 - (t+20) = 3$ $t^2 + 22t + 121 - t -

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Given equation: $(x^2-9x+11)^2-(x^2-9x+20)=3$ Let $t = x^2-9x$. Substituting $t$ into the equation: $(t+11)^2 - (t+20) = 3$ $t^2 + 22t + 121 - t - 20 - 3 = 0$ $t^2 + 21t + 98 = 0$ $(t+14)(t+7) = 0$ So,$t = -7$ or $t = -14$. Case $1$: $x^2-9x = -7 \Rightarrow x^2-9x+7 = 0$. The roots are $x = \frac{9 \pm \sqrt{81-28}}{2} = \frac{9 \pm \sqrt{53}}{2}$ (irrational). Case $2$: $x^2-9x = -14 \Rightarrow x^2-9x+14 = 0$. $(x-7)(x-2) = 0 \Rightarrow x = 7, 2$ (rational). The rational roots are $7$ and $2$. The product of the rational roots is $7 	imes 2 = 14$.

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

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