Q: Find the values of ‘a ‘ for which the equation
(x2 + x + 2)2 – (a – 3) (x2 + x + 2) (x2 + x + 1) +( a – 4) (x2 + x +1)2 = 0 has at least one real root.
Sol: Putting x2 + x +1 = α ,
we get (α +1)2 – ( a – 3)α( α +1) + ( a – 4) α2 = 0
⇒ α (5 – a ) + 1 = 0
⇒ α =-1/(5-a)
⇒ x2 + x + 1 =1/(a-5)
$\displaystyle \frac{1}{a-5} \ge \frac{3}{4}$
⇒ a – 5 > 0 and a – 5 ≤ 4/3
⇒ 5 < a ≤ 19/3.