"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"

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Accepted Answer

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To solve the given equation, start by rewriting the expression:

$ (x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3 $

First, simplify the second part of the expression:

$ (x - 4)(x - 5) = x^2 - 9x + 20 $

Now the equation becomes:

$ (x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3 $

Introduce a substitution for simplification:

$ \text{Let } t = x^2 - 9x $

Thus, the equation transforms to:

$ t^2 + 22t + 121 - t - 20 - 3 = 0 $

Simplify further:

$ t^2 + 21t + 98 = 0 $

Factor the quadratic:

$ (t + 14)(t + 7) = 0 $

This gives:

$ t = -7 \quad \text{or} \quad t = -14 $

Address each case where $ t = x^2 - 9x $:

$ x^2 - 9x = -7 $

$ x^2 - 9x + 7 = 0 $

Solving this quadratic equation, we find the roots:

$ x = \frac{9 \pm \sqrt{81 - 4 \times 7}}{2} = \frac{9 \pm \sqrt{53}}{2} $

$ x^2 - 9x = -14 $

$ x^2 - 9x + 14 = 0 $

Solving this quadratic equation:

$ x = \frac{9 \pm \sqrt{81 - 4 \times 14}}{2} = \frac{9 \pm \sqrt{25}}{2} $

$ x = \frac{9 \pm 5}{2} = 7 \quad \text{or} \quad x = 2 $

The rational roots from the second equation are 7 and 2. Thus, the product of all the rational roots is:

$ 7 \times 2 = 14 $

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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