For t>-1, let αₜ and βₜ be the roots of the equation
((t+2)^(1 / 7)-1) x²+((t+2)^(1 / 6)-1) x+((t+2)^(1 / 21)-1)=0 . If lim _(t →-1⁺) αₜ=a and lim _(t →-1⁺) βₜ=b ,
then 72(a+b)² is equal to ___________.

Solution

Maths sir · 4 steps

Every coefficient tends to 0, so divide the whole equation by t + 1 before taking the limit.

1. Zero by zero form. Put s = t + 2. As t → -1⁺, s → 1⁺, so each coefficient sᵖ - 1 tends to 0. The sum of roots, -(s^(1/6)-1)/(s^(1/7)-1), is then a 0/0 form.
2. Scale the equation. Since s > 1, s - 1 ≠ 0, so dividing the equation by s - 1 does not change its roots. Each coefficient becomes (sᵖ-1)/(s-1), which tends to p.

   lim(s → 1) (sᵖ-1)/(s-1) = p ⇒ 1/7x² + 1/6x + 1/21 = 0

3. Limiting equation. Multiply by 42 to get 6x² + 7x + 2 = 0, that is, (2x + 1)(3x + 2) = 0. So a and b are -1/2 and -2/3, and a + b = -7/6.
4. So. 72(a+b)² = 72 × 49/36 = 2 × 49 = 98.

The common slip: 8. Using the product of roots instead of the sum gives 72 × (1/3)² = 8, not 98.

The idea to keep. When every coefficient of an equation tends to 0, divide them all by the same small quantity first. Here (sᵖ-1)/(s-1) → p turns each coefficient into its exponent.

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