Question

If one root of $x^2+px-q^2=0$, p and q are real, be less than 2, and other be greater than 2. Then

• A. 4+2p+q2>0
• B. 4+2p+q2<0
• C. 4+2p-q2>0
• D. 4+2p-q2<0

Answer 1

Answer: (D)

4+2p-q2<0

Answer 2

1. The given quadratic equation is x^2 + px - q^2 = 0.
2. We know that the sum of the roots of a quadratic equation is -p, and the product of the roots is -q^2 (from Vieta's formulas).
3. We're given that one root is less than 2, and the other is greater than 2.
4. For the product of the roots to be negative (one root less than 2, and the other greater than 2), their signs must be opposite.
5. Let the roots be a and b. Without loss of generality, assume that a < 2 and b > 2.
6. This implies that a * b < 0.
7. From Vieta's formulas, we have a * b = -q^2, which means that q^2 > 0.
8. Rewriting the equation: q^2 < 0 is not possible (since q^2 > 0), so q^2 > 0.
9. From the original equation, p = -(a + b).
10. Since one root is less than 2 and the other is greater than 2, we can write: a < 2 and b > 2, which means -(a + b) < 0, so p < 0.
11. Putting p and q^2 together: 4 + 2p - q^2 < 0.

The correct answer is option (D): 4+2p-q2<0