The product of all the rational roots of the equation (x²-9 x+11)²-(x-4)(x-5)=3, is equal to

Solution

Maths sir · 4 steps

Spot that (x-4)(x-5) is x²-9x+20, so one substitution turns this quartic into a quadratic.

1. Find the common block. Expand (x-4)(x-5) = x²-9x+20. Put t = x²-9x+11, so (x-4)(x-5) = t + 9.

   t² - (t+9) = 3 ⇒ t² - t - 12 = 0 ⇒ (t-4)(t+3) = 0

2. Case t = 4. x²-9x+11 = 4 gives x²-9x+7 = 0. Its discriminant is 81-28 = 53, which is not a perfect square, so both roots are irrational.
3. Case t = −3. x²-9x+11 = -3 gives x²-9x+14 = 0, that is (x-2)(x-7) = 0. Both roots, 2 and 7, are rational.
4. So. The rational roots are 2 and 7, and their product is 2 × 7 = 14, option (d).

The idea to keep. When the same expression in x keeps showing up, call it t and solve the smaller equation first. Then check each case with the discriminant to see which roots are rational.

Answer from the NTA final answer key. Solution by PrepWiser.