Let the set of all values of p ∈ ℝ, for which both the roots of the equation x²-(p+2) x+(2 p+9)=0 are negative real numbers, be the interval (α, β]. Then β-2 α is equal to

Solution

Maths sir · 4 steps

Both roots negative means three checks: real roots, negative sum, positive product. Intersect them.

1. Three conditions. Both roots are real and negative exactly when D ≥ 0, the sum of the roots is negative, and the product of the roots is positive. Here the sum is p+2 and the product is 2p+9.

   D = (p+2)² - 4(2p+9) = p² - 4p - 32 = (p-8)(p+4)

2. Solve each one. D ≥ 0 gives p ≤ -4 or p ≥ 8. The sum condition p+2<0 gives p<-2. The product condition 2p+9>0 gives p>-9/2.
3. Intersect. All three hold together only for -9/2 < p ≤ -4. So α = -9/2 and β = -4.
4. So. β - 2α = -4 + 9 = 5, which is option (a).

The idea to keep. For both roots negative, use D ≥ 0, sum < 0 and product > 0. For both positive, flip only the sum condition. Always intersect all three.

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