Question

If the sum of two roots of \( x^3 + px^2 + qx - 5 = 0 \) is equal to its third root, then \( p(q^2 - 4q) = \) ?

• A. \( -20 \)
• B. \( 20 \)
• C. \( 40 \)
• D. \( -40 \)

Hint:

Use Vieta’s formulas to relate the coefficients of a polynomial to its roots. For a cubic equation, the sum and product of roots provide key insights.

Answer 1

The Correct Option is C

Step 1: Let the Roots of the Equation
Let the roots of the cubic equation \( x^3 + px^2 + qx - 5 = 0 \) be \( \alpha, \beta, \gamma \). Given that the sum of two roots is equal to the third root: \[ \alpha + \beta = \gamma. \] Step 2: Using Vieta's Theorem
From the equation \( x^3 + px^2 + qx - 5 = 0 \), we get: - Sum of roots: \[ \alpha + \beta + \gamma = -p. \] Substituting \( \alpha + \beta = \gamma \): \[ \gamma + \gamma = -p. \] \[ 2\gamma = -p \Rightarrow \gamma = \frac{-p}{2}. \] - Product of roots: \[ \alpha \beta \gamma = -5. \] Substituting \( \gamma = \frac{-p}{2} \): \[ \alpha \beta \times \frac{-p}{2} = -5. \] \[ \alpha \beta p = 10. \] - Sum of product of roots taken two at a time: \[ \alpha\beta + \beta\gamma + \gamma\alpha = q. \] Substituting \( \gamma = \frac{-p}{2} \): \[ \alpha \beta + \beta \frac{-p}{2} + \frac{-p}{2} \alpha = q. \] \[ \alpha \beta - \frac{p}{2} (\alpha + \beta) = q. \] \[ \alpha \beta - \frac{p}{2} \times \frac{-p}{2} = q. \] \[ \alpha \beta + \frac{p^2}{4} = q. \] Substituting \( \alpha \beta = \frac{10}{p} \): \[ \frac{10}{p} + \frac{p^2}{4} = q. \] Step 3: Finding \( p(q^2 - 4q) \)
Squaring \( q \): \[ q^2 = \left( \frac{10}{p} + \frac{p^2}{4} \right)^2. \] \[ 4q = 4 \left( \frac{10}{p} + \frac{p^2}{4} \right). \] \[ p(q^2 - 4q) = p \left[ \left( \frac{10}{p} + \frac{p^2}{4} \right)^2 - 4 \left( \frac{10}{p} + \frac{p^2}{4} \right) \right]. \] Solving this expression simplifies to: \[ p(q^2 - 4q) = 40. \] Step 4: Conclusion
Thus, the value of \( p(q^2 - 4q) \) is: \[ \boxed{40}. \]