Find the Domain of the Square Root of a Quadratic Function

For a function \( f(x) = \sqrt{ax^2 + bx + c} \) to have real values, the radicand (expression under the square root) must satisfy:

\( ax^2 + bx + c \ge 0 \)

The domain is the set of all \( x \) that satisfy this quadratic inequality. The solution depends on the sign of \( a \) and the discriminant \( \Delta = b^2 - 4ac \).

📌 The graph below shows the function \( f(x) \) (blue) and its radicand (green parabola). The function exists only where the parabola is above or on the x-axis.

✧ Domain of √(ax² + bx + c) ✧

Enter coefficients a, b, c of the quadratic expression

Quadratic Expression: \( ax^2 + bx + c \)

\( f(x) = \sqrt{ \; ax^2 + bx + c \; } \)   →   \( f(x) = \sqrt{x^2 - 4x + 3} \)

⚠️ a cannot be zero (otherwise it's not quadratic). If a=0, it will be set to 1.
📐 Domain of \( f(x) = \sqrt{ax^2 + bx + c} \)
\( (-\infty, 1] \cup [3, \infty) \)

📖 Step-by-Step Solution

STEP 1: Set the radicand greater than or equal to zero
\( x^2 - 4x + 3 \ge 0 \)
STEP 2: Solve the quadratic equation \( ax^2 + bx + c = 0 \) and determine the inequality solution
STEP 3: Write the domain in interval notation
\( (-\infty, 1] \cup [3, \infty) \)
💡 Graphical Interpretation
The blue curve is \( f(x) = \sqrt{ax^2+bx+c} \), the green parabola is \( y = ax^2+bx+c \). The function exists only where the green parabola is on or above the x-axis.
📊 Interactive Graph
f(x) = √(ax²+bx+c) y = ax²+bx+c (radicand) Domain boundary points
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