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

Question

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

Accepted Answer

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:

$t={x}^{2}-9x$

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:

$t=-7\, \, t=-14$

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:

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

${x}^{2}-9x=-14$

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

Solving this quadratic equation:

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

$x=\frac{9\pm 5}{2}=7\, \, x=2$

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

$7	imes 2=14$

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