Question

If \( a \) and \( b \) are the non-zero distinct roots of \( x^2 + ax + b = 0 \), then the minimum value of \( x^2 + ax + b \) is:

• A. \( \frac{2}{3} \)
• B. \( \frac{9}{4} \)
• C. \( \frac{-9}{4} \)
• D. \( \frac{-2}{3} \)
• E. \( 1 \)

Hint:

The vertex of any quadratic \( y = Ax^2 + Bx + C \) occurs at \( x = -B/2A \). Substituting this \( x \) back into the original equation always yields the minimum (if \( A>0 \)) or maximum (if \( A<0 \)) value.

Answer 1

The Correct Option is C

Concept: For a quadratic equation \( x^2 + ax + b = 0 \) with roots \( a \) and \( b \): 1. Sum of roots: \( a + b = -a \implies 2a + b = 0 \) 2. Product of roots: \( a \cdot b = b \implies a = 1 \) (since \( b \neq 0 \)) After finding \( a \) and \( b \), the minimum value of the quadratic \( f(x) = x^2 + ax + b \) is given by the formula \( \frac{-D}{4A} \), where \( D = a^2 - 4b \).

Step 1: Find the values of \( a \) and \( b \).
From the product of roots: \( ab = b \implies a = 1 \). From the sum of roots: \( a + b = -a \implies 1 + b = -1 \implies b = -2 \). The expression becomes: \( f(x) = x^2 + x - 2 \).

Step 2: Find the minimum value of the quadratic.
The minimum value occurs at \( x = -a/2 = -1/2 \). \[ f(-1/2) = (-1/2)^2 + (-1/2) - 2 \] \[ f(-1/2) = \frac{1}{4} - \frac{1}{2} - 2 = \frac{1 - 2 - 8}{4} = -\frac{9}{4} \]