Question

If \(\alpha,\beta\) are the rational roots and \(l,m\) are the irrational roots of \[ (x^2-9x+11)^2-(x-4)(x-5)=3, \] then \(\alpha+\beta+lm=\)

• A. \(2\)
• B. \(16\)
• C. \(8\)
• D. \(12\)

Hint:

In higher degree equations, always search for repeated quadratic expressions. Substitution often reduces the equation dramatically.

Answer 1

The Correct Option is B

Concept: Whenever a quartic equation contains a perfect square expression, try to simplify by substitution or factorization.

Step 1: Bring all terms to one side.
\[ (x^2-9x+11)^2-(x-4)(x-5)-3=0 \] Since \[ (x-4)(x-5)+3 = x^2-9x+20+3 = x^2-9x+23 \] Let \[ t=x^2-9x+11 \] Then \[ t^2-(t+12)=0 \] \[ t^2-t-12=0 \] \[ (t-4)(t+3)=0 \]

Step 2: Find corresponding quadratic equations.
Case 1: \[ t=4 \] \[ x^2-9x+7=0 \] Roots: \[ x=\frac{9\pm\sqrt{53}}{2} \] These are irrational. Hence \[ lm=\frac{7}{1}=7 \] Case 2: \[ t=-3 \] \[ x^2-9x+14=0 \] \[ (x-7)(x-2)=0 \] Thus \[ \alpha=7,\qquad \beta=2 \] \[ \alpha+\beta=9 \]

Step 3: Compute the required value.
\[ \alpha+\beta+lm = 9+7 = 16 \] \[ \boxed{16} \]