Question

for which of the following conditions does $ax^2 + bx + c = 0$ have two real solutions?

$\bigcirc$ $b^2 = 4ac$

$\bigcirc$ $b^2 > 4ac$

$\bigcirc$ $a = b, \\ c = b$

Explanation:

Step1: Recall Discriminant Rule

The discriminant of quadratic \(ax^2 + bx + c = 0\) is \(D = b^2 - 4ac\).

• If \(D > 0\) (i.e., \(b^2 - 4ac > 0\) or \(b^2 > 4ac\)), two distinct real solutions.
• If \(D = 0\) (i.e., \(b^2 = 4ac\)), one real solution (a repeated root).
• If \(D < 0\), no real solutions.

Step2: Analyze Each Option

• Option 1 (\(b^2 = 4ac\)): \(D = 0\), one real solution (not two).
• Option 2 (\(b^2 > 4ac\)): \(D > 0\), two distinct real solutions.
• Option 3 (\(a = b, c = b\)): Substituting, \(D = b^2 - 4b \cdot b = -3b^2\). If \(b

eq 0\), \(D < 0\) (no real solutions); if \(b = 0\), \(D = 0\) (one solution). Not two real solutions.

Answer:

\( \boldsymbol{b^2 > 4ac} \) (the second option)