Question

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

• A. 14
• B. 7
• C. 28
• D. 21

Hint:

When solving higher degree polynomial equations, use the Rational Root Theorem to list all possible rational roots. Test each possible root by substituting it into the equation and using synthetic division if necessary.

Answer 1

The Correct Option is A

Step 1: Expand and simplify the equation. We are given the equation \( (x^2 - 9x + 11)^2 - (x-4)(x-5) = 3 \). 
First, let's expand the terms: \[ (x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3. \] \[ (x^2 - 9x + 11)^2 - (x^2 - 9x + 20) - 3 = 0. \] 
Step 2: Use a substitution to simplify the equation. Let \( y = x^2 - 9x + 11 \). Then \( x^2 - 9x = y - 11 \). 
The equation becomes: \[ y^2 - (y - 11 + 20) - 3 = 0. \] \[ y^2 - (y + 9) - 3 = 0. \] \[ y^2 - y - 12 = 0. \] 
Step 3: Solve the quadratic equation for \( y \). Factoring the quadratic gives \[ (y-4)(y+3) = 0. \] So \( y = 4 \) or \( y = -3 \). 
Step 4: Substitute back to find the values of \( x \).
Case 1: \( y = 4 \). Then \( x^2 - 9x + 11 = 4 \implies x^2 - 9x + 7 = 0 \). 
The roots are \( x = \frac{9 \pm \sqrt{81 - 28}}{2} = \frac{9 \pm \sqrt{53}}{2} \), which are irrational.
Case 2: \( y = -3 \). Then \( x^2 - 9x + 11 = -3 \implies x^2 - 9x + 14 = 0 \). 
The roots are \( x = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2} \). 
So \( x = \frac{14}{2} = 7 \) or \( x = \frac{4}{2} = 2 \).
Step 5: Find the product of the rational roots. The rational roots are 7 and 2. Their product is \( 7 \times 2 = 14 \). 
Final Answer: The product of all the rational roots is 14.