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

The Correct Option is D

1. The given quadratic equation is: $x^2 + px - q^2 = 0$.
2. Sum of roots = $-p$, and Product of roots = $-q^2$ (by Vieta’s formulas). 
3. We’re told that one root is less than 2 and the other is greater than 2.
4. This means the roots are on either side of 2, so the function changes sign at $x = 2$.
5. Let the roots be $a$ and $b$, such that $a < 2 < b$.
6. Then their product $ab = -q^2 < 0$ implies the roots have opposite signs.
7. Since $q^2 > 0$, this is valid.
8. Also, the sum $a + b > 4$ (since both are around 2), so $p = -(a + b) < -4$ → p is negative.
9. Now, the expression to evaluate is: $4 + 2p - q^2$.
10. Since $p < 0$ and $q^2 > 0$, this expression is less than zero.

Therefore, the correct condition is: $4 + 2p - q^2 < 0$

The correct option is (D)

Answer 2

Step 1: Understand the Condition

For a quadratic \(ax^2 + bx + c\) to have one root less than \(k\) and another root greater than \(k\), we require that \(a \cdot p(k) < 0\). In this case, \(a = 1\), and \(k = 2\).

Therefore, we need \(p(2) < 0\).

Step 2: Evaluate p(2)

\(p(2) = (2)^2 + p(2) - q^2 = 4 + 2p - q^2\)

Step 3: Apply the Condition

Since \(p(2) < 0\), we have:

\(4 + 2p - q^2 < 0\)

Conclusion:

The condition on \(p\) and \(q\) is:

\(4 + 2p - q^2 < 0\)