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

To solve this equation, we start by setting up the given expression: 

\(\left((x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3\right)\)

Step 1: Simplify the expression:

First, calculate \((x - 4)(x - 5)\):

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

Step 2: Substitute this into the original equation:

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

Step 3: Simplify the equation:

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

Step 4: Let \(y = x^2 - 9x + 11\). Then the equation becomes:

\(y^2 - y = 3 + x^2 - 9x - 20\)

Substitute \(y\) back:

\(y^2 - y = -17\)

Solve this quadratic equation:

\(y^2 - y + 17 = 0\)

Solve for \(y\) using the quadratic formula:

The roots are complex, and the rational roots condition suggests we look for:

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

Step 5: Returning to the quadratic, equate degrees:

Check the likely integer roots using factorizing techniques or trial-and-error method:

Using either approach:

The roots of the polynomial give the rational solutions:

\(x = 7, 2\) (Here calculated through trial.)

Step 6: Multiply these roots to find:

\(7 \times 2 = 14\)

Thus, the product of all rational roots is 14.

Answer 2

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.