Digital SAT Math

SAT Math: In 3x² - 12x + c = 0, If No Real Solutions, Find Least Integer Value of c

2026-08-18 4 min read
• Direct Answer / Key Takeaway

For 3x^2 - 12x + c = 0 to have no real solutions, the discriminant must be strictly negative: b^2 - 4ac < 0. Here, (-12)^2 - 4(3)(c) < 0 => 144 - 12c < 0 => c > 12. The least possible integer value of c is 13.

Exact Question Text

In the given equation, $c$ is a constant:
$$3x^2 - 12x + c = 0$$
If the equation has no real solutions, what is the least possible integer value of $c$?

Step-by-Step Algebraic Solution

  1. Identify the Quadratic Coefficients:

In $ax^2 + bx + c = 0$: $a = 3$, $b = -12$, and constant term $= c$.

  1. Apply the Discriminant Condition:

The number of real solutions is governed by $\Delta = b^2 - 4ac$: $$\Delta < 0 \implies \text{no real solutions}$$

  1. Set up the Inequality:

$$(-12)^2 - 4(3)(c) < 0$$ $$144 - 12c < 0 \implies c > 12$$

  1. Determine the Least Integer:

Since $c > 12$, the smallest integer is $13$.


Fast Desmos 10-Second Shortcut

  1. In Line 1, type: y = 3x^2 - 12x + c and add a slider for $c$.
  2. At $c = 12$, the parabola touches the axis at $(2, 0)$.
  3. For $c \ge 13$, the curve floats with 0 real roots. Answer: 13.

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