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SAT discriminant: how many solutions, without solving

"The equation has exactly one solution. What is the value of c?" You started factoring. Factoring can get there when the quadratic happens to be a clean perfect square — but it is a search, and the discriminant is a calculation.

Why it happens

Questions that say "exactly one", "no real", or "two distinct" are counting solutions, not finding them. The count is decided entirely by b² − 4ac: positive gives two, zero gives one, negative gives none. Solving instead turns a one-line calculation into an algebra problem carrying an unknown constant — and when the numbers are not friendly, into guesswork.

See the trap

x² + 6x + c = 0 has exactly one real solution. Find c.

Factoring into (x + ?)(x + ?) and guessing values of c that work.

Exactly one solution means b² − 4ac = 0, so 36 − 4c = 0 and c = 9.

How to actually fix it

  1. 1Read the wording first: "exactly one" → discriminant = 0, "no real" → < 0, "two distinct" → > 0.
  2. 2Put the equation in ax² + bx + c = 0 form before reading off a, b and c — a term left on the wrong side is the usual sign error.
  3. 3Substitute into b² − 4ac and solve that, not the original equation.

Practice this trap · 8 real questions

Each of these is a question from the free bank that turns on the mistake above — the trap is the reason its wrong answer looks right. Easiest first.

FAQ

What if the question gives a system with a line and a parabola?

Set them equal, collect everything on one side, and it becomes a single quadratic. "Intersect at exactly one point" then means the same thing as "exactly one solution", and the discriminant answers it the same way.

How do I stop repeating this mistake?

Reading the explanation isn't enough — your brain needs to beat the trap on a fresh question. Practice the skill in the kaoshen.co question bank: miss one and it is logged with the exact error, then brought back to you three days later and again after fourteen.

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